14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence T=
1 1 = = 2.27 × 10 −3 s = 2.27 ms f 440 Hz
14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your heart’s oscillations is thus 75 beats = 1.25 beats s = 1.25 Hz 60 s The period is the inverse of the frequency, hence 1 1 T= = = 0.80 s f 1.25 Hz f =
14.3. Model: The air-track glider oscillating on a spring is in simple harmonic motion. Solve:
The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.
(a)
T=
33 s = 3.3 s oscillation = 3.3 s 10 oscillations
(b)
f =
1 1 = = 0.303 Hz T 3.3 s
(c)
ω = 2πf = 2π (0.303 Hz) = 1.904 rad s
(d) The oscillation from one side to the other is equal to 60 cm – 10 cm = 50 cm = 0.50 m. Thus, the amplitude is A = 12 (0.50 m ) = 0.25 m. (e) The maximum speed is 2π vmax = ωA = A = (1.904 rad s)(0.25 m ) = 0.476 m s T
14.4. Model: The air-track glider attached to a spring is in simple harmonic motion.
Visualize: The position of the glider can be represented as x(t) = A cos ω t. Solve: The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period T = 2.0 s and a maximum speed vmax = 40 cm s = 0.40 m s . (a)
vmax = ωA and ω =
v 2π 2π 0.40 m s = = π rad s ⇒ A = max = = 0.127 m = 12.7 cm T ω π rad s 2.0 s
(b) The glider’s position at t = 0.25 s is x 0.25 s = (0.127 m ) cos (π rad s)(0.25 s) = 0.090 m = 9.0 cm
[
]
14.5. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.5. Solve: (a) The amplitude A = 10 cm. (b) The time to complete one cycle is the period, hence T = 2.0 s and 1 1 f = = = 0.50 Hz T 2.0 s (c) The position of an object undergoing simple harmonic motion is x (t ) = A cos(ω t + φ0 ) . At t = 0 s, x0 = 5 cm, thus
5 cm = (10 cm ) cos[ω (0 s) + φ 0 ] ⇒ cos φ 0 =
π 5 cm 1 1 = ⇒ φ 0 = cos −1 = rad or 60° 10 cm 2 2 3
14.6. Model: The oscillation is the result of simple harmonic motion. Visualize: Please refer to Figure Ex14.6. Solve: (a) The amplitude A = 20 cm. (b) The period T = 4.0 s, thus
1 1 = = 0.25 Hz T 4.0 s (c) The position of an object undergoing simple harmonic motion is x (t ) = A cos(ω t + φ0 ) . At t = 0 s, x 0 = −10 cm. Thus, 10 cm 1 2π rad = ±120° −10 cm = (20 cm ) cos φ 0 ⇒ φ 0 = cos −1 − = cos −1 − = ± 20 cm 2 3 Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thus must have a phase constant between π and 2π radians. Therefore, φ 0 = − 23 π rad = −120° . f =
π has a plus sign, which implies that the object undergoing simple harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left. Solve: The position of the object is
14.7. Visualize: The phase constant
2 3
[
x(t ) = A cos(ωt + φ 0 ) = A cos(2πft + φ 0 ) = ( 4.0 cm ) cos ( 4π rad s)t + 23 π rad
]
The amplitude is A = 4 cm and the period is T = 1 f = 0.50 s. A phase constant φ 0 = 2π 3 rad = 120° (second quadrant) means that x starts at − 12 A and is moving to the left (getting more negative).
Assess: We can see from the graph that the object starts out moving to the left.
14.8. Visualize: A phase constant of − 23 π implies that the object that undergoes simple harmonic motion is in the third quadrant of the circular motion diagram. That is, the object is moving to the right. Solve: The position of the object is given by the equation
[
x(t ) = A cos(ωt + φ 0 ) = A cos(2πft + φ 0 ) = (6.0 cm ) cos (π rad s)t − 23 π rad
]
The amplitude is A = 6 cm and the period is T = 1 f = 2.0 s. With φ 0 = −2π 3 rad, x starts at − 12 A and is moving to the right (getting more positive).
Assess: As we see from the graph, the object starts out moving to the right.
14.9. Solve: The position of the object is given by the equation x(t ) = A cos(ωt + φ 0 ) = A cos(2πft + φ 0 ) We can find the phase constant φ 0 from the initial condition:
0 cm = ( 4.0 cm ) cos φ 0 ⇒ cos φ 0 = 0 ⇒ φ 0 = cos −1 (0) = ± 12 π rad Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence, φ 0 = − 12 π rad. The final result, with f = 4.0 Hz, is
[
x(t ) = ( 4.0 cm ) cos (8.0π rad s)t − 12 π rad
]
14.10. Solve: The position of the object is given by the equation x(t ) = A cos(ωt + φ 0 ) = A cos(2πft + φ 0 ) We can find the phase constant φ 0 from the initial condition:
− A = A cos φ 0 ⇒ φ 0 = cos −1 ( −1) = π rad The final result, with f = 0.50 Hz, is
[
x(t ) = (10.0 cm ) cos (π rad s)t + π
]
14.11. Model: The air-track glider is in simple harmonic motion. Solve:
(a) We can find the phase constant from the initial conditions for position and velocity: x0 = Acosφ0
v0x = –ωAsinφ0
Dividing the second by the first, we see that
sin φ 0 v = tan φ 0 = − 0 x cos φ 0 ωx 0 The glider starts to the left (x0 = –5.0 cm) and is moving to the right (v0x = +36.3 cm/s). With a period of 1.5 s = 3 4 2 s, the angular frequency is ω = 2π/T = 3 π rad/s. Thus
1 36.3 cm / s 2 φ 0 = tan −1 − = 3 π rad (60°) or – 3 π rad (–120°) (4π / 3 rad / s) ( −5.0 cm) The tangent function repeats every 180°, so there are always two possible values when evaluating the arctan function. We can distinguish between them because an object with a negative position but moving to the right is in the third quadrant of the corresponding circular motion. Thus φ0 = − 23 π rad or –120°. (b) At time t, the phase is φ = ω t + φ 0 = ( 43 π rad / s) t − 23 π rad . This gives φ = − 23 π rad, 0 rad, 23 π rad, and 4 3 π rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
14.12. Model: The block attached to the spring is in simple harmonic motion. Solve:
The period of an object attached to a spring is
T = 2π
m = T0 = 2.0 s k
where m is the mass and k is the spring constant. (a) For mass = 2m,
T = 2π (b) For mass
1 2
2m = k
( 2 )T
0
= 2.83 s
m,
T = 2π
1 2
m = T0 k
2 = 1.41 s
(c) The period is independent of amplitude. Thus T = T0 = 2.00 s (d) For a spring constant = 2k,
T = 2π
m = T0 2k
2 = 1.41 s
14.13. Model: The air-track glider attached to a spring is in simple harmonic motion. Solve:
Experimentally, the period is T = (12.0 s) (10 oscillations) = 1.20 s. Using the formula for the period,
m 2π 2π ⇒k= m= (0.200 kg) = 5.48 N m T 1.20 s k 2
T = 2π
2
14.14. Model: The oscillating mass is in simple harmonic motion. Solve: (a) The amplitude A = 2.0 cm. (b) The period is calculated as follows:
ω=
2π 2π = 10 rad s ⇒ T = = 0.628 s T 10 rad s
(c) The spring constant is calculated as follows:
ω=
k 2 ⇒ k = mω 2 = (0.050 kg)(10 rad s) = 5.0 N m m
(d) The phase constant φ 0 = − 14 π rad. (e) The initial conditions are obtained from the equations
x(t ) = (2.0 cm ) cos(10t − 14 π ) and v x (t ) = −(20.0 cm s) sin(10t − 14 π ) At t = 0 s, these equations become
x 0 = (2.0 cm ) cos( − 14 π ) = 1.414 cm and v0 x = −(20.0 cm s) sin( − 14 π ) = 14.14 cm s In other words, the mass is at +1.414 cm and moving to the right with a velocity of 14.14 cm/s. (f) The maximum speed is vmax = Aω = (2.0 cm )(10 rad s) = 20.0 cm s . 2 (g) The total energy E = 12 kA 2 = 12 (5.0 N m )(0.02 m ) = 1.0 × 10 −3 J . (h) At t = 0.40 s, the velocity is
[
]
v0 x = −(20.0 cm s) sin (10 rad s)(0.40 s) − 14 π = 1.46 cm s
14.15. Model: The mass attached to the spring oscillates in simple harmonic motion. Solve: (a) The period T = 1 f = 1 2.0 Hz = 0.50 s . (b) The angular frequency ω = 2πf = 2π (2 Hz) = 4π rad /s . (c) Using energy conservation 1 2
kA 2 = 12 kx 02 + 12 mv02x
Using x0 = 5.0 cm, v0x = −30 cm/s and k = mω2 = (0.2 kg) (4π rad s) 2 , we get A = 5.54 cm. (d) To calculate the phase constant φ 0 , A cos φ 0 = x 0 = 5.0 cm
⇒ φ 0 = cos −1
5.0 cm = 0.445 rad 5.54 cm
(e) The maximum speed is vmax = ωA = ( 4π rad s)(5.54 cm ) = 69.6 cm s . (f) The maximum acceleration is amax = ω 2 A = ω (ωA) = ( 4π rad s)(69.6 cm s) = 875 cm s 2 2 (g) The total energy is E = 12 mvmax = (h) The position at t = 0.40 s is
1 2
(0.200 kg)(0.696 m s)2 = 0.0484 J .
[
]
x 0.4 s = (5.54 cm ) cos ( 4π rad s)(0.40 s) + 0.445 rad = +3.81 cm
14.16. Model: The mass attached to the spring is in simple harmonic motion. Solve:
(a) The period is
T = 2π
m = 2π k
(0.507 kg) (20 N m)
= 1.00 s
(b) The angular frequency is ω = 2π T = 2π 1.0 s = 2π rad s. (c) To calculate the phase constant φ 0 ,
x 0 = A cos φ 0 ⇒ 0.05 m = (0.10 m ) cos φ 0 ⇒ φ 0 = cos −1 ( 12 ) = ± 13 π rad Because the mass is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram. Hence, φ 0 = − 13 π rad. (d) At t = 0 s, v x (t ) = −(0.10 m )(2π rad s) sin(2πt + φ 0 ) becomes
−(0.10 m )(2π rad s) sin( − 13 π rad ) = 0.544 m s
(e) The maximum speed vmax = ωA = (2π rad s)(0.10 m ) = 0.628 m s .
(g) At t = 1.3 s, x1.3 s = (0.10 m ) cos[2π (1.3 s) − 13 π rad ] = 0.0669 m .
[
]
(h) At t = 1.3 s, v(1.3 s ) x = −(0.10 m )(2π rad s) sin (2π rad s)(1.3 s) − 13 π rad = −0.467 m s .
14.17. Model: The block attached to the spring is in simple harmonic motion. Visualize:
Solve:
(a) The conservation of mechanical energy equation Kf + Usf = Ki + Usi is 1 2
mv12 + 12 k ( ∆x ) = 12 mv02 + 0 J ⇒ 0 J + 12 kA 2 = 12 mv02 + 0 J 2
m 1.0 kg v0 = (0.40 m s) = 0.10 m = 10.0 cm k 16 N m (b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equation K 2 + Us2 = K i + Usi is ⇒A=
2
1 2 1 A 1 1 1 1 1 1 1 1 3 1 mv2 + k = mv02 + 0 J ⇒ mv22 = mv02 − kA 2 = mv02 − mv02 = mv02 2 42 2 2 2 2 2 2 42 42 ⇒ v2 =
3 v0 = 4
3 (0.40 m s) = 0.346 m s = 34.6 cm s 4
14.18. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
(a) At equilibrium, Newton’s first law applied to the physics book is
(F )
sp y
− mg = 0 N ⇒ − k∆y − mg = 0 N
⇒ k = − mg ∆y = − (0.500 kg)(9.8 m s 2 ) ( −0.20 m ) = 24.5 N m (b) To calculate the period:
ω=
k = m
24.5 N m 2π 2π rad = 7.0 rad s and T = = = 0.898 s 0.50 kg ω 7.0 rad s
(c) The maximum speed is
vmax = Aω = (0.10 m )(7.0 rad s) = 0.70 m s Maximum speed occurs as the book passes through the equilibrium position.
14.19. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
The period and angular frequency are
T=
20 s 2π 2π = 0.6667 s and ω = = = 9.425 rad s 30 oscillations T 0.6667 s
(a) The mass can be found as follows:
ω=
k k 15.0 N / m ⇒m= 2 = = 0.169 kg m ω (9.425 rad s)2
(b) The maximum speed vmax = ωA = (9.425 rad s)(0.060 m ) = 0.565 m/s.
14.20. Model: The vertical oscillations constitute simple harmonic motion. Solve: To find the oscillation frequency using ω = 2πf = k m , we first need to find the spring constant k. In equilibrium, the weight mg of the block and the spring force k∆L are equal and opposite. That is, mg = k∆L ⇒ k = mg ∆L . The frequency of oscillation f is thus given as
f =
1 2π
k 1 = m 2π
mg ∆L 1 = m 2π
g 1 = ∆L 2π
9.8 m s 2 = 3.52 Hz 0.020 m
14.21. Solve:
(a) and (b)
θ (°)
θ (rad)
sin θ
cos θ
0
0
0
1.0000
2
.0349
.0349
.9994
4
.0698
.0698
.9976
6
.1047
.1045
.9945
8
.1396
.1392
.9903
10
.1745
.1736
.9848
12
.2094
.2079
.9781
(c) 12° (d) 10° (e) The small-angle approximation is good to three significant figures for θ up to 10°.
14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:
The period of the pendulum is
T0 = 2Ď€
L0 = 4.0 s g
(a) The period is independent of the mass and depends only on the length. Thus T = T0 = 4.0 s. (b) For a new length L = 2L0, 2 L0 T = 2Ď€ = 2 T0 = 5.66 s g (c) For a new length L = L0/2,
L0 2 1 = T0 = 2.83 s g 2 (d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4.0 s. T = 2Ď€
14.23. Model: Assume the small-angle approximation so there is simple harmonic motion. Solve:
The period is T = 12.0 s 10 oscillations = 1.20 s and is given by the formula 2
T = 2π
2
L T 1.20 s ⇒L= g= 9.8 m s 2 ) = 35.7 cm 2π 2π ( g
14.24. Model: Assume a small angle of oscillation so there is simple harmonic motion. Solve:
(a) On the earth the period is
Tearth = 2π
L 1.0 m = 2π = 2.01 s g 9.80 m s 2
(b) On Venus the acceleration due to gravity is
gVenus =
−11 24 2 2 GMVenus (6.67 × 10 N ⋅ m kg )( 4.88 × 10 kg) = 8.86 m s 2 = 2 2 6 RVenus (6.06 × 10 m)
⇒ TVenus = 2π
L gVenus
= 2π
1.0 m = 2.11 s 8.86 m s 2
14.25. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoes simple harmonic motion. Solve: Using the formula g = GM R 2 , the periods of the pendulums on the moon and on the earth are
Tearth = 2π
2 2 Learth Rearth Lmoon Rmoon L = 2π and Tmoon = 2π g GMearth GMmoon
Because Tearth = Tmoon, 2
2π
2 2 M R Learth Rearth Lmoon Rmoon = 2π ⇒ Lmoon = moon earth Learth GMearth GM moon Mearth Rmoon 2
7.36 × 10 22 kg 6.37 × 10 6 m = (2.0 m ) = 33 cm 5.98 × 10 24 kg 1.74 × 10 6 m
14.26. Model: The pendulum undergoes simple harmonic motion. Solve: (a) The amplitude is 0.10 rad. (b) The frequency of oscillations is
f = (c) The phase constant φ = π rad. (d) The length can be obtained from the period:
ω 5 = Hz = 0.796 Hz 2π 2π
2
2
1 g 1 2 ω = 2πf = ⇒L= g= (9.8 m s ) = 0.392 m L 2 π 0.796 Hz 2πf ( ) (e) At t = 0 s, θ 0 = (0.10 rad ) cos(π ) = −0.10 rad . To find the initial condition for the angular velocity we take the derivative of the angular position: dθ ( t ) θ (t ) = (0.10 rad ) cos(5t + π ) ⇒ = −(0.10 rad )(5) sin(5t + π ) dt At t = 0 s, ( dθ dt ) 0 = ( −0.50 rad ) sin(π ) = 0 rad s .
(f) At t = 2.0 s, θ 2.0 = (0.10 rad ) cos(5(2.0 s) + π ) = 0.084 rad.
14.27. Model: The motion is a damped oscillation.
Solve: The amplitude of the oscillation at time t is given by Equation 14.55: A(t ) = A0 e − t 2τ , where τ = m b is the time constant. Using x = 0.368 A and t = 10.0 s, we get
0.368 A = Ae −10.0 s 2τ ⇒ ln(0.368) =
−10 s 10.0 s ⇒τ = − = 5.00 s 2τ 2ln(0.368)
14.28. Model: The motion is a damped oscillation.
Solve: The position of a damped oscillator is x (t ) = Ae ( ) cos(ωt + φ 0 ) . The frequency is 1.0 Hz and the damping time constant τ is 4.0 s. Let us assume φ 0 = 0 rad and A = 1 with arbitrary units. Thus, − t 2τ
x(t ) = e − t (8.0 s ) cos[2π (1.0 Hz)t ] ⇒ x(t ) = e −0.125 t cos(2πt ) where t is in s. Values of x(t) at selected values of t are displayed in the following table: t(s)
x(t)
t(s)
x(t)
t(s)
x(t)
0
1
2.00
0.779
6.00
0.472
0.25
0
2.50
– 0.732
6.50
– 0.444
0.50
– 0.939
3.00
0.687
7.00
0.417
0.75
0
3.50
– 0.646
7.50
– 0.392
1.00
0.882
4.00
0.607
8.00
0.368
1.25
0
4.50
– 0.570
8.50
– 0.346
1.50
– 0.829
5.00
0.535
9.00
0.325
1.75
0
5.50
– 0.503
9.50
– 0.305
10.00
0.286
14.29. Model: The motion is a damped oscillation. Solve:
The position of the air-track glider is x (t ) = Ae
− ( t 2τ )
cos(ωt + φ 0 ) , where τ = m b and
k b2 − m 4m 2
ω= Using A = 0.20 m, φ 0 = 0 rad , and b = 0.015 kg/s,
4.0 N m (0.015 kg s) 16 − 9 × 10 −4 rad s = 4.00 rad s − 2 = 0.250 kg 4(0.250 kg) 2
ω= Thus the period is
2π 2π rad = = 1.57 s ω 4.0 rad s The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e −1 A at a time given by 1 m − t 2τ A = Ae ( ) ⇒ t = 2τ = 2 = 33.3 s b e The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21. T=
14.30. Model: The vertical oscillations are damped and follow simple harmonic motion. Solve: The position of the ball is given by x (t ) = Ae of time. The angular frequency is
− ( t 2τ )
cos(ωt + φ 0 ) . The amplitude A(t ) = Ae
− ( t 2τ )
is a function
(15.0 N m) = 5.477 rad s ⇒ T = 2π = 1.147 s k = m ω 0.500 kg Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s = 34.41 s, we have −34.41 s − 34.414 s 2 τ ) − 34.41 s 2 τ ) 3.0 cm = (6.0 cm )e ( ⇒ 0.5 = e ( ⇒ ln(0.5) = ⇒ τ = 24.8 s 2τ ω=
14.31. Model: The object is undergoing simple harmonic motion. Visualize:
Solve:
The formula for the period is
1 m = 2Ď€ f k (a) When the frequency f is halved, the period is doubled. That is, the period increases from 4.0 s to 8.0 s. (b) When the mass m is quadrupled, the period doubles. T=
14.32. Model: The object is undergoing simple harmonic motion. Visualize:
Solve:
The velocity of an object oscillating on a spring is
v x (t ) = − Aω sin(ωt + φ 0 ) (a) For A ′ = 2 A and ω ′ = ω 2 , we have
[
]
v x′ (t ) = −(2 A)(ω 2) sin (ω 2)t + φ 0 = − Aω sin(ωt 2 + φ 0 ) That is, the maximum velocity A ′ω ′ remains the same, but the frequency of oscillation is halved. (b) For m ′ = 4 m ,
ω′ =
k = m′
ω k = and A ′ = A ⇒ vmax ′ = A ′ω ′ = A ω 2 = vmax 2 4m 2
Quadrupling of the mass halves the frequency or doubles the time period, and halves the maximum velocity.
14.33. Visualize: Please refer to Figure P14.33. Solve:
The position and the velocity of a particle in simple harmonic motion are
x(t ) = A cos(ωt + φ 0 ) and v x (t ) = − Aω sin(ωt + φ 0 ) = − vmax sin(ωt + φ 0 ) (a) At t = 0 s, the equation for x yields
(5.0 cm) = (10.0 cm) cos(φ 0 ) ⇒ φ 0 = cos −1 (0.5) = ± 13 π rad Because the particle is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram, and the phase constant is between π and 2π radians. Thus, φ 0 = − 13 π rad. (b) At t = 0 s,
π 2π v0 x = − Aω sin φ 0 = −(10.0 cm ) sin − = +6.80 cm T 3 (c) The maximum speed is
2π vmax = ωA = (10.0 cm) = 7.85 cm s 8.0 s Assess: The positive velocity at t = 0 s is consistent with the position-versus-time graph and the negative sign of the phase constant.
14.34. Visualize: Please refer to Figure P14.34. Solve:
The position and the velocity of a particle in simple harmonic motion are
x(t ) = A cos(ωt + φ 0 ) and v x (t ) = − Aω sin(ωt + φ 0 ) = − vmax sin(ωt + φ 0 ) From the graph, T = 12 s and the angular frequency is
ω=
2π 2π π rad s = = T 12 s 6
(a) Because vmax = Aω = 60 cm s , we have
A= (b) At t = 0 s,
60 cm s 60 cm s = = 114.6 cm ω π 6 rad s
v0 x = − Aω sin φ 0 = −30 cm ⇒ −(60 cm s) sin φ 0 = −30 cm ⇒ φ 0 = sin −1 (0.5 rad ) = 16 π rad (30°) or 65 π rad (150°)
Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant of the circular motion diagram. Thus φ 0 = 65 π rad. (c) At t = 0 s, x 0 = (114.6 cm ) cos( 65 π rad ) = −99.2 cm .
14.35. Model: The vertical mass/spring systems are in simple harmonic motion. Visualize: Please refer to Figure P14.35. Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to the maximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz. (b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first time is at t = 1.5 s. The time period of system B is thus 6.0 s. (c) Spring/mass A undergoes three oscillations in 12 s, giving it a period TA = 4.0 s. Spring/mass B undergoes 2 oscillations in 12 s, giving it a period TB = 6.0 s. We have
TA = 2π
m k 4.0 s 2 mA mB T and TB = 2π ⇒ A = A B = = kA kB TB mB kA 6.0 s 3
If mA = mB, then
kB 4 k 9 = ⇒ A = = 2.25 kA 9 kB 4
14.36. Model: The astronaut attached to the spring is in simple harmonic motion. Visualize: Please refer to Figure P14.36. Solve: (a) From the graph, T = 3.0 s, so we have 2
T = 2π
2
m T 3.0 s ⇒m= k= (240 N m) = 54.7 kg 2π 2π k
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A = 0 rad, and
ω=
1 2
(0.80 m) = 0.40 m, φ 0 =
2π 2π = = 2.094 rad s T 3.0 s
The equation for the position of the astronaut is
[
]
x(t ) = A cos ωt + 1.0 m = (0.4 m ) cos (2.094 rad s)t + 1.0 m
[
]
[
]
⇒ 1.2 m = (0.4 m ) cos (2.094 rad s)t + 1.0 m ⇒ cos (2.094 rad s)t = 0.5 ⇒ t = 0.50 s The equation for the velocity of the astronaut is
v x (t ) = − Aω sin(ωt )
[
]
⇒ v0.5 s = −(0.4 m )(2.094 rad s) sin (2.094 rad s)(0.50 s) = −0.725 m s Thus her speed is 0.725 m/s.
14.37. Model: The block attached to the spring is in simple harmonic motion. Visualize:
The position and the velocity of the block are given by the equations
x(t ) = A cos(ωt + φ 0 ) and v x (t ) = − Aω sin(ωt + φ 0 ) Solve: To graph x(t) we need to determine ω, φ 0 , and A. These quantities will be found by using the initial (t = 0 s) conditions on x(t) and vx(t). The period is
T = 2π
m 2π 1.0 kg 2π rad = 2π = 1.405 s ⇒ ω = = = 4.472 rad s k T 20 N m 1.405 s
At t = 0 s, x 0 = A cos φ 0 and v0 x = − Aω sin φ 0 . Dividing these equations,
tan φ 0 = −
v0 x (−1.0 m s) =− = 1.1181 ⇒ φ 0 = 0.841 rad ωx 0 (4.472 rad s)(0.20 m)
From the initial conditions, 2
A=
x 02 +
v0 x = ω
2
−1.0 m s = 0.300 m 4.472 rad s
(0.20 m)2 +
The position-versus-time graph can now be plotted using the equation
[
x(t ) = (0.300 m ) cos ( 4.472 rad s)t + 0.841 rad
]
14.38. Solve: The object’s position as a function of time is x(t ) = A cos(ωt + φ 0 ) . Letting x = 0 m at t = 0 s, gives 0 = Acos φ 0 ⇒ φ 0 = ± 12 π Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant between −π and 0 radians. Thus, φ 0 = − 12 π and
x(t ) = A cos(ωt − 12 π ) ⇒ x(t ) = A sin ωt = (0.10 m ) sin( 12 πt ) where we have used A = 0.10 m and
ω=
2π 2π rad π = = rad s T 4.0 s 2
Let us now find t where x = 0.060 m:
2 0.060 m π = 0.410 s 0.060 m = (0.10 m ) sin t ⇒ t = sin −1 2 0.10 m π Assess: The answer is reasonable because it is approximately
1 8
of the period.
14.39. Model: The particle is in simple harmonic motion. Solve:
The equation for the velocity of the particle is v x (t ) = −(25 cm )(10 rad s) sin(10 t )
Substituting into K = 2U gives 1 2 1 1 mv x (t ) = 2 kx 2 (t ) ⇒ m −(250 cm s) sin(10 t ) 2 2 2
[
]
2
= k[(25 cm ) cos(10 t )]
sin 2 (10 t ) k 1 2 (25 cm) = 2 = 2ω 2 s 100 m (250 cm s) 2 cos 2 (10 t ) 2
⇒
1 2 1 2 tan −1 2.0 = 0.0955 s ⇒ tan 2 (10 t ) = 2(10 rad s) s = 2.0 ⇒ t = 100 10
2
14.40. Model: The spring undergoes simple harmonic motion. Solve: (a) Total energy is E =
1 2
kA2. When the displacement is x =
U = 12 kx = 12 k ( 12 A) = 2
2
1 4
(
1 2
kA
2
)=
1 4
1 2
A, the potential energy is
E ⇒ K = E − U = 43 E
One quarter of the energy is potential and three-quarters is kinetic. (b) To have U = 12 E requires
U = 12 kx 2 = 12 E =
1 2
(
1 2
kA 2 ) ⇒ x =
A 2
14.41. Solve: Average speed is vavg = ∆x/∆t. During half a period (∆t = x = +A (∆x = 2A). Thus
vavg =
1 2
T), the particle moves from x = –A to
∆x 2A 4A 4A 2 2 π = = = = (ω A) = vmax ⇒ vmax = vavg ∆t T /2 T 2π / ω π 2 π
14.42. Model: The block on a spring is in simple harmonic motion.
Solve: (a) The position of the block is given by x (t ) = A cos(ωt + φ 0 ) . Because x (t ) = A at t = 0 s, we have φ 0 = 0 rad , and the position equation becomes x(t ) = A cos ωt . At t = 0.685 s, 3.0 cm = A cos(0.685ω ) and at t = 0.886 s, −3.0 cm = A cos(0.886ω ) . These two equations give
cos(0.685ω ) = − cos(0.886ω ) = cos(π − 0.886ω ) ⇒ 0.685ω = π − 0.886ω ⇒ ω = 2.00 rad s (b) Substituting into the position equation,
3.0 cm = A cos((2.00 rad s)(0.685 s)) = A cos(1.370) = 0.20 A ⇒ A =
3.0 cm = 15.0 cm 0.20
14.43. Model: The ball attached to a spring is in simple harmonic motion. Solve:
(a) Let t = 0 s be the instant when x0 = −5 cm and v0 = 20 cm/s. The oscillation frequency is
k = m
ω=
2.5 N m = 5.0 rad / s 0.10 kg
Using Equation 14.27, the amplitude of the oscillation is 2
A=
x 02 +
v0 = ω
2
20 cm / s = 6.40 cm 5 rad / s
(−5 cm)2 +
(b) The maximum acceleration is amax = ω 2 A = 160 cm s 2 . (c) For an oscillator, the acceleration is most positive
(a = amax )
when the displacement is most negative
( x = − x max = − A) . So the acceleration is maximum when x = −6.40 cm . (d) We can use the conservation of energy between x 0 = −5 cm and x1 = 3 cm: 1 2
mv02 + 12 kx 02 = 12 mv12 + 12 kx12 ⇒ v1 = v02 +
k 2 ( x0 − x12 ) = 0.283 m s = 28.3 cm s m
Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.
14.44. Solve: (a) The velocity is –0.25 m/s at times that satisfy the equation −0.25 m / s = ( −0.35 m / s) sin(20t + π ) ⇒ sin(20t + π ) =
−0.25 m / s = 0.7143 −0.35 m / s
The time is
20t + π rad = sin −1 (0.7143) = 0.7956 rad ⇒ t =
0.7956 rad − π rad = −0.117 s 20 rad / s
This is a time at which v = –0.25 m/s, but it’s not a time after t = 0 s. But the sine function is periodic with period 2π rad, so another angle whose sine is 0.7143 is
20t + π rad = sin −1 (0.7143) = 0.7956 rad + 2π rad ⇒ t =
0.7956 rad + π rad = 0.197 s 20 rad / s
This is a time t > 0 s at which v = –0.25 m/s. (b) The amplitude is A = vmax/ω = (0.35 m/s)(20 rad/s) = 0.0175 m = 1.75 cm. Thus the position at t = 0.197 s is
x = A cos(ωt + φ 0 ) = (1.75 cm) cos((20 rad / s) (0 / 197 s) + π rad) = 1.22 cm
14.45. Model: The oscillator is in simple harmonic motion. Energy is conserved. Solve:
The energy conservation equation E1 = E2 is 1 2
mv12 + 12 kx12 = 12 mv22 + 12 kx 22
1 1 1 1 (0.30 kg)(0.954 m s)2 + k(0.030 m)2 = (0.30 kg)(0.714 m s)2 + k(0.060 m)2 2 2 2 2 ⇒ k = 44.48 N m The total energy of the oscillator is
Etotal =
1 2 1 2 1 1 2 2 mv1 + kx1 = (0.30 kg)(0.954 m s) + ( 44.48 N m )(0.030 m ) = 0.1565 J 2 2 2 2
2 Because Etotal = 12 mvmax ,
0.1565 J =
1 2 ⇒ v max = 1.02 m s (0.300 kg)vmax 2
Assess: A maximum speed of 1.02 m/s is reasonable.
14.46. Model: The transducer undergoes simple harmonic motion. Solve:
Newton’s second law for the transducer is
Frestoring = ma max ⇒ 40, 000 N = (0.10 × 10 −3 kg)amax ⇒ amax = 4.0 × 10 8 m s 2 Because amax = ω 2 A ,
A=
amax 4.0 × 10 8 m s 2 = 2 ω 2π (1.0 × 10 6 Hz)
[
2
= 1.01 × 10 −5 m = 10.1 µm
(
)
]
(b) The maximum velocity is
vmax = ωA = 2π (1.0 × 10 6 Hz) 1.01 × 10 −5 m = 63.6 m s
14.47. Model: The block attached to the spring is in simple harmonic motion. Solve:
(a) The frequency is
f =
1 2π
k 1 = m 2π
2000 N m = 3.183 Hz 5.0 kg
(b) From energy conservation, 2
A=
x 02 +
2
1.0 m / s v0 = 0.0707 m = (0.050 m ) 2 + 2π ⋅ 3.183 Hz ω
(c) The total mechanical energy is
E = 12 kA 2 =
1 2
(2000 N m)(0.0707 m)2 = 5.0 J
14.48. Model: The spider is in simple harmonic motion.
Solve: Your tapping is a driving frequency. Largest amplitude at fext = 1.0 Hz means that this is the resonance frequency, so f0 = fext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f0 is 1.0 Hz and ω 0 = 2πf0 = 2π rad s . We have
ω0 =
k 2 ⇒ k = mω 02 = (0.002 kg)(2π rad s) = 0.079 N m m
14.49. Model: The block undergoes simple harmonic motion. Visualize:
Solve:
(a) The frequency of oscillation is
f = (b) Using conservation of energy,
x1 =
x 02 +
1 2
1 2π
k 1 = m 2π
10 N / m = 1.125 Hz 0.20 kg
mv12 + 12 kx12 = 12 mv02 + 12 kx 02 , we find
m 2 0.20 kg (v0 − v12 ) = ( −0.20 m) 2 + ((1.00 m / s) 2 − (0.50 m / s) 2 ) k 10.0 N / m
= 0.235 m or 23.5 cm (c) At time t, the displacement is x = Acos(ω t + φ0). The angular frequency is ω = 2π f = 7.071 rad/s. The amplitude is 2
A=
x 02 +
2
1.00 m / s v0 = ( −0.20 m) 2 + = 0.245 m ω 7.071 rad / s
The phase constant is
x −0.200 m φ 0 = cos −1 0 = cos −1 = ±2.526 rad or ± 145° 0.245 m A A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that the corresponding circular motion is in the third quadrant, so φ0 = –2.526 rad. Thus at t = 1.0 s,
x = (0.245 m) cos((7.071 rad / s) (1.0 s) − 2.526 rad ) = −0.409 m = −4.09 cm The block is 4.09 cm below the equilibrium point.
14.50. Model: The mass is in simple harmonic motion. Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation. Gravitational potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches the spring, then the elastic potential energy is converted 100% back into gravitational potential energy as the mass rises, bringing the mass back to exactly its starting height. The total displacement of the oscillation – high point to low point – is 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can deduce that the equilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about this equilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm below the equilibrium point. Solve: The equilibrium point is the point where the mass would hang at rest, with Fsp = w = mg. At the equilibrium point, the spring is stretched by ∆y = 10 cm = 0.1 m. Hooke’s law is Fsp = k∆y , so the equilibrium condition is
[F
sp
]
= k∆y = [w = mg] ⇒
k g 9.8 m s 2 = = = 98 s −2 m ∆y 0.1 m
The ratio k m is all we need to find the oscillation frequency:
f =
1 2π
k 1 = 98 s −2 = 1.58 Hz m 2π
14.51. Model: The compact car is in simple harmonic motion. Solve:
(a) The mass on each spring is (1200 kg) 4 = 300 kg . The spring constant can be calculated as follows:
2 k 2 ⇒ k = mω 2 = m(2πf ) = (300 kg)[2π (2.0 Hz)] = 4.74 × 10 4 N m m (b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is, m = 300 kg + 70 kg = 370 kg . Thus,
ω2 =
f =
ω 1 = 2π 2π
k 1 = m 2π
4.74 × 10 4 N m = 1.80 Hz 370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the square root of the mass.
14.52. Model: The blocks undergo simple harmonic motion. Visualize:
The length of the stretched spring due to a block of mass m is ∆L1 . In the case of the two block system, the spring is further stretched by an amount ∆L2 . Solve: The equilibrium equations from Newton’s second law for the single block and double block systems are
(∆L1 )k = mg and (∆L1 + ∆L2 )k = (2m)g Using ∆L2 = 5.0 cm , and subtracting these two equations, gives us
(∆L1 + ∆L2 )k − ∆L1k = (2m)g − mg ⇒ (0.05 m)k = mg
⇒
ω k 9.8 m s 2 = = ω 2 ⇒ ω = 14.0 rad s ⇒ f = = 2.23 Hz 2π m 0.05 m
14.53. Model: Hooke’s law for the spring. The spring’s compression and decompression constitutes simple harmonic motion. Visualize:
Solve: (a) The spring’s compression or decompression is one-half of the oscillation cycle. This means the contact time is ∆t = 12 T , where T is the period. The period is calculated as follows:
ω=
k = m
1 2π 2π 50 N m = 10 rad s ⇒ T = = = = 0.628 s 0.50 kg ω 10 rad s f
T = 0.314 s 2 (b) There is no change in contact time, because period of oscillation is independent of the amplitude or the maximum speed. ⇒ ∆t =
14.54. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply the model of static friction between the two blocks. Visualize:
Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2. The model of static friction gives the maximum force of static friction as
fs max = µ s n = µ s ( m1 g) = m1amax ⇒ amax = µ s g
(
)
Using µ s = 0.5 , amax = µ S g = (0.5) 9.8 m s 2 = 4.9 m s 2 . That is, the two blocks will ride together if the maximum acceleration of the system is equal to or less than amax. We can calculate the maximum value of A as follows:
amax = ω 2 Amax =
2 a ( m + m2 ) ( 4.9 m s )(1.0 kg + 5.0 kg) k Amax ⇒ Amax = max 1 = = 0.588 m 50 N m m1 + m2 k
14.55. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. We will also use the model of static friction between the two blocks. Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2. The two blocks ride together as long as the static friction doesn’t exceed its maximum possible value. The model of static friction gives the maximum force of static friction as fs max = µ s n = µ s ( m1 g) = m1amax ⇒ amax = µ s g
⇒ µs = Assess:
2 2 amax ω 2 Amax 2π Amax 2π 0.40 m = = = = 0.716 T g 1.5 s 9.8 m s 2 g g
Because the period is given, we did not need to use the block masses or the spring constant in our calculation.
14.56. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simple harmonic motion. Visualize:
Solve:
(a) Using the formula for the period of a pendulum, 2
T = 2π
2
L T 5.5 s = 7.51 m ⇒ L = g = (9.8 m s 2 ) 2π 2π g
(b) The conservation of mechanical energy equation K 0 + U g0 = K1 + U g1 for the swinging lamp is 1 2
2 mv02 + mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mgh = 12 mvmax +0 J
⇒ vmax = 2 gh = 2 g( L − L cos 3°) = 2(9.8 m s 2 )(7.51 m )(1 − cos 3°) = 0.449 m s
14.57. Model: Assume that the pendulum makes a small angle with the vertical so that there is simple harmonic motion. Solve: The angle θ made by the string with the vertical as a function of time is
θ (t ) = θ max cos(ωt + φ 0 ) The pendulum starts from maximum displacement, thus φ 0 = 0 . Thus, θ (t ) = θ max cos ωt . To find the time t when the pendulum reaches 4.0° on the opposite side:
(−4.0°) = (8.0°) cosωt ⇒ ωt = cos −1 (−0.5) = 2.094 rad Using the formula for the angular frequency,
ω=
g = L
2.0944 rad 2.094 rad 9.8 m s 2 = 3.130 rad s ⇒ t = = = 0.669 s 3.130 rad s 1.0 m ω
Assess: Because T = 2π ω = 2.00 s , a value of 0.669 s for the pendulum to cover a little less than half the oscillation is reasonable.
14.58. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion. Solve:
The time periods of the pendulums on the earth and on Mars are
Tearth = 2π
L L and TMars = 2π gearth g Mars
Dividing these two equations,
Tearth = TMars
2
2 T gMars 1.50 s ⇒ gMars = gearth earth = (9.8 m s 2 ) = 3.67 m s 2 2.45 s gearth TMars
14.59. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion. Solve:
(a) At the equator, the period of the pendulum is
Tequator = 2π
1.000 m = 2.009 s 9.78 m s 2
The time for 100 oscillations is 200.9 s. (b) At the north pole, the period is
Tpole = 2π
1.000 m = 2.004 s 9.83 m s 2
The time for 100 oscillations is 200.4 s. (c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operated stopwatch. (d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated by rearranging the formula for the period: 2
gmountain
2 2π 2π = 9.77 m s 2 = L = (1.000 m ) 2.010 s Tmountain
Assess: This last result is reasonable because g decreases with altitude.
14.60. Model: The pendulum is a damped oscillator. Solve:
The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
T = 2π
4(3600 s) L 15.0 m = 2π = 7.773 s ⇒ N osc = = 1853 g 9.8 m s 2 7.773 s
The amplitude of the pendulum as a function of time is A(t ) = Ae − bt /2 m . The exponent of this expression can be calculated to be
−
(0.010 kg s)(4 × 3600 s) = −0.6545 bt =− 2m 2(110 kg)
We have A(t ) = (1.5 m )e −0.6545 = 0.780 m.
14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion. Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2, which have initial velocities vi1 and vi2. After m1 collides with and sticks to m2, the two masses move together with velocity vf. Solve: The momentum conservation equation pf = pi for the completely inelastic collision is ( m1 + m2 )vf = m1vi1 + m2 vi2 . Substituting the given values,
(0.750 kg)vf = (0.25 kg)(1.20 m s) + (0.50 kg)(0 m s) ⇒ vf
= 0.40 m s
We now use the conservation of mechanical energy equation:
( K + Us )compressed = ( K + Us )equilibrium ⇒ 0 J + 12 kA2 = 12 (m1 + m2 )vf2 + 0 J ⇒A=
m1 + m2 vf = k
0.750 kg (0.40 m s) = 0.110 m 10 N m
The period is
T = 2π
m1 + m2 0.750 kg = 2π = 1.72 s 10 N m k
14.62. Model: The block attached to the spring is oscillating in simple harmonic motion. Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/or the maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz. (b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanical energy equation as follows: 1 2
⇒ v0 =
2 kA 2 = 12 mvmax = 12 mv02
k A = ωA = (2πf ) A = (2π )(2.0 Hz)(0.02 m ) = 0.25 m s m
The blow to the block provides an impulse that changes the velocity of the block:
J x = Fx ∆t = ∆p = mvf − mv0
(−20 N)(1.0 × 10 −3 s) = (0.200 kg)vf − (0.200 kg)(0.25 m s) ⇒ vf = 0.15 m s Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω. Thus,
A=
0.15 m s 0.15 m s = = 0.012 m = 1.2 cm 2π (2.0 Hz) ω
Assess: Because vf is positive, the block continues to move to the right even after the blow.
14.63. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion. Visualize:
Solve:
(a) The equation for conservation of energy after the collision is
1 2 1 kA = ( mb + mB )vf2 ⇒ vf = 2 2
k A= m b + mB
2500 N m (0.10 m) = 4.975 m s 1.010 kg
The momentum conservation equation for the perfectly inelastic collision pafter = pbefore is
( m b + m B )v f
= m b v b + mB vB
(1.010 kg)(4.975 m s) = (0.010 kg)vb + (1.00 kg)(0 m s) ⇒ vb = 502
ms
(b) No. The oscillation frequency k ( mb + mB ) depends on the masses but not on the speeds.
14.64. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion. Visualize:
We placed the origin of the coordinate system at the bottom of the arc. Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for the = K + Ug : pendulum’s fall is K + U g
(
)
1 2
top
(
)
bottom
mv02 + mgy0 = 12 mv12 + mgy1 ⇒ 0 J + mg(2 L) = 12 m(5.0 m s) + 0 J 2
1 (5.0 m s) = 0.6377 m 4 g 2
⇒L=
Using L = 0.6377 m, we can find the frequency f as
f =
1 2π
1 g = L 2π
9.8 m s 2 = 0.624 Hz 0.6377 m
14.65. Model: Assume the small-angle approximation. Visualize:
Solve:
The tension in the two strings pulls downward at angle θ. Thus Newton’s second law is
∑ Fy = −2T sin θ = may From the geometry of the figure we can see that
sin θ =
y L + y2 2
If the oscillation is small, then y << L and we can approximate sinθ ≈ y/L. Since y/L is tanθ, this approximation is equivalent to the small-angle approximation sinθ ≈ tanθ if θ << 1 rad. With this approximation, Newton’s second law becomes
−2T sin θ ≈ −
2T d2y y = ma y = m 2 L dt
⇒
d2y 2T =− y dt 2 mL
This is the equation of motion for simple harmonic motion (see Eqs. 14.33 and 14.47). The constants 2T/mL are equivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is
f =
1 2π
2T mL
14.66. Visualize: Please refer to Figure P14.66.
Solve: The potential energy curve of a simple harmonic oscillator is described by U = 12 k ( ∆x ) , where ∆x = x − x 0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 = 0.13 nm. We can find the bond’s spring constant by reading the value of the potential energy U at a displacement ∆x and using the potential energy formula to calculate k. 2
∆x (nm)
x (nm)
0.02
0.11
U (J)
k (N/m)
0.8 × 10
−19
J
400
−19
0.10
0.03
1.9 × 10
J
422
0.09
0.04
3.4 × 10−19 J
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the spring constant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
f =
1 2π
k 1 = m 2π
416 N m = 7.9 × 1013 Hz 1.67 × 10 −27 kg
14.67. Model: Assume that rmarble << Rhoop and that θ is a small angle. Visualize:
Solve:
The marble is like an object on an inclined plane. The net force on the marble in the tangential direction is
− w sin θ = ma = mRα = mR
d 2θ d 2θ ⇒ − mg sin θ = mR 2 2 dt dt
where α is the angular acceleration. With the small-angle approximation sin θ ≈ θ , this becomes
d 2θ g = − θ = −ω 2θ dt 2 R This is the equation of motion of an object in simple harmonic motion with a period of
T=
2π R = 2π g ω
14.68.
Visualize:
Solve:
(a) Newton’s second law applied to the penny along the y-axis is
Fnet = n − mg = ma y r Fnet is upward at the bottom of the cycle (positive ay), so n > mg. The speed is maximum when passing through r equilibrium, but ay = 0 so n = mg. The critical point is the highest point. Fnet points down and ay is negative. If ay becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface. (b) When the penny loses contact (n = 0), the equation for Newton’s law becomes amax = g . For simple harmonic motion,
g 9.8 m s 2 = = 15.65 rad A 0.04 m ω 15.65 rad s ⇒ f = = = 2.49 Hz 2π 2π
amax = Aω 2 ⇒ ω =
14.69. Model: The vertical oscillations constitute simple harmonic motion. Visualize:
Solve:
At the equilibrium position, the net force on mass m on Planet X is:
Fnet = k∆L − mgX = 0 N ⇒
g k = X m ∆L
For simple harmonic motion k m = ω 2 , thus
ω2 =
gX ⇒ω = ∆L
2
2 2π gX 2π 2π 2 = ⇒ gX = ∆L = (0.312 m ) = 5.86 m s T ∆L T 14.5 s 10
14.70. Model: The doll’s head is in simple harmonic motion and is damped. Solve:
(a) The oscillation frequency is
f =
1 2π
k 2 2 2 ⇒ k = m(2πf ) = (0.015 kg)(2π ) ( 4.0 Hz) = 9.475 N m m
(b) The maximum speed is
vmax = ωA =
k A= m
9.475 N m (0.020 m) = 0.503 m s 0.015 kg
(c) Using A(t ) = A0 e − bt / 2 m , we get
(0.5 cm) = (2.0 cm)e − b( 4.0 s ) ( 2×0.015 kg ) ⇒ 0.25 = e − (133.33 s kg )b ⇒ −(133.33 s kg)b = ln 0.25 ⇒ b = 0.0104 kg / s
14.71. Model: The oscillator is in simple harmonic motion. Solve:
(a) The maximum displacement at time t of a damped oscillator is
x max (t ) = Ae − t 2τ ⇒ −
t x (t ) = ln max A 2τ
Using xmax = 0.98A at t = 0.5 s, we can find the time constant τ to be 0.5 s τ =− = 12.375 s 2ln(0.98) 25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be
x max, 12.5 s = (10.0 cm )e −12.5 s ( 2 )(12.375 s ) = 6.03 cm (b) The energy of a damped oscillator decays more rapidly than the amplitude: E(t) = E 0e–t/τ. When the energy is 60% of its initial value, E(t)/E0 = 0.60. We can find the time this occurs as follows:
−
E (t ) E (t ) t = ln ⇒ t = −τ ln = −(12.375 s) ln(0.60) = 6.32 s τ E0 E0
14.72. Model: The oscillator is in simple harmonic motion.
Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as xmax(t) = Ae − t 2τ , where τ is the time constant. We know x max A = 0.60 at t = 50 s, so we can find τ as follows:
t 50 s x (t ) = ln max ⇒ τ = − = 48.9 s A 2τ 2 ln(0.60) Now we can find the time t30 at which x max A = 0.30 : −
x (t ) t30 = −2τ ln max = −2( 48.9 s) ln(0.30) = 118 s A The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillation frequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, which allows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Then the number of oscillations before the spring decays to 30% of its initial amplitude is
N = f ⋅ t30 = (2 oscillations s) ⋅ (118 s) = 236 oscillations
14.73. Model: Samson on the trapeze is in simple harmonic motion. Visualize:
The time (t1D − t 0D ) for Delilah to fall 4.0 m must be equal to the time (t1S − t 0S ) Samson takes to swing from the angle θ to the lowest point of the arc. Solve: The time of Delilah’s fall can be found from kinematics as follows:
y1 − y0 = v0y (t1D − t 0D ) + 12 ay (t1D − t 0D ) 0 m − 4.0 m = (0 m s)(t1D − t 0D ) +
1 2
(−9.8 m s )(t 2
1D
2
− 0 s) ⇒ t1D = 0.9035 s 2
Treating Samson as a pendulum, his period is
L 6.0 m = 2π = 4.916 s g 9.8 m s 2 If Samson starts his swing at t = 0 s, his motion is θ = θmaxcos(ω t) = (8.0°)cos[(2π/4.916 s)t)]. He reaches the lowest point (θ = 0°) at t1S = 14 T = 1.229 s. In order for Samson and Delilah to meet at this point, she had to jump 0.9035 s earlier, at t0S = 0.3255 s. Samson’s angle at that instant was T = 2π
2π θ 0 = (8.0°)cos (0.3255 s) = 7.3° 4.916 s Delilah should jump when Samson makes a 7.3° angle on his downward swing.
14.74. Model: The astronaut will be attached to a spring and will oscillate in simple harmonic motion. She can determine her mass by measuring her period. Solve: (a)
(b) One end of a spring is attached to the wall of the space ship. The spring was calibrated prior to leaving earth and its spring constant was measured. An astronaut is strapped to the other end of the spring. Another astronaut pulls her back and releases her. She will oscillate back and forth on the spring with a period T = 2π m k . It is easy to measure the period accurately with a stopwatch. Her mass is determined from the above equation. (c) The only required component is the spring. The design value of the spring constant is 300 N/m. The actual value will be determined by calibration measurements before launch. (d) The period needs to be slow enough to be easily measured (and also slow enough not to make the astronaut sick!) But it can’t be so slow the measurements take too long. A period of approximately 3 s is reasonable, allowing the astronaut to oscillate 10 times in approximately 30 s. Astronaut masses are most likely to be in the range 50 kg (100 pounds) to 100 kg (220 pounds). Sample oscillation periods for three different astronauts are Mass m
Period T for k = 300 N/m
50 kg
2.57 s
75 kg
3.14 s
100 kg
3.63 s
A target value of k ≈ 300 N/m gives acceptable oscillation periods for the full range of possible astronaut masses. Assess: Spring constants ranging from k ≈ 30 N/m (T = 10 s for 75 kg) to k ≈ 700 N/m (T = 2 s for 75 kg) would be acceptable designs.
14.75. Model: The two springs obey Hooke’s law. Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces – one pushing, the other pulling – are directed to the left and have negative values:
( Fnet ) x = ( Fsp 1 ) x + ( Fsp 2 ) x = − k1 x − k2 x = −(k1 + k2 ) x = − keff x
where keff = k1 + k2 is the effective spring constant. This means the oscillatory motion of the block under the influence of the two springs will be the same as if the block was attached to a single spring with spring constant keff. The frequency of the blocks, therefore, is
f =
1 2π
keff 1 = m 2π
k1 + k2 = m
k1 k2 + = 4π 2 m 4π 2 m
f12 + f22
14.76. Model: The two springs obey Hooke’s law. Assume massless springs. Visualize:
Each spring is shown separately. Note that ∆x = ∆x1 + ∆x2.
Solve: Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m. Newton’s third law tells us that F2 on m = Fm on 2 and that F2 on 1 = F1 on 2. From Fnet = ma, the net force on a massless spring is zero. Thus Fw on 1 = F2 on 1 = k1∆x1 and Fm on 2 = F1 on 2 = k2∆x2. Combining these pieces of information, Fm = k1∆x1 = k2∆x2 The net displacement of the mass is ∆x = ∆x1 + ∆x2, so
∆x = ∆x1 + ∆x 2 =
Fm Fm 1 k + k2 1 + = + Fm = 1 Fm k1 k2 k1 k2 k1k2
Turning this around, the net force on the mass is kk kk Fm = 1 2 ∆x = keff ∆x where keff = 1 2 k1 + k2 k1 + k2 keff, the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring constant. Thus the mass’s angular frequency of oscillation is
ω=
keff = m
1 k1k2 m k1 + k2
Using ω 12 = k1 / m and ω 22 = k2 / m for the angular frequencies of either spring acting alone on m, we have
ω=
( k1 / m) ( k2 / m) = ( k1 / m) + ( k2 / m)
ω 12ω 22 ω 12 + ω 22
Since the actual frequency f is simply a multiple of ω, this same relationship holds for f:
f =
f 12 f 22 f + f 22 2 1
14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the motion. Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward. Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit on the spring at rest. The spring is compressed by ∆L at this point. Balancing the forces requires k∆L = mg. The angular frequency is w 2 = k/m = g/∆L, so we can find the oscillation frequency by finding ∆L. The block hits the spring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have been transformed into the spring’s elastic energy. Equate the energies at these points: K1 + U1g = U3s + U3g ⇒ 12 mv12 + mg∆L = 12 k ( ∆L + A) 2 + mg( − A) We’ve used y1 = ∆L as the block hits and y3 = –A at the bottom. The spring has been compressed by ∆y = ∆L + A. Speed v1 is the speed after falling distance h, which from free-fall kinematics is v12 = 2gh. Substitute this expression for v12 and mg/∆L for k, giving
mgh + mg∆L =
mg ( ∆L + A) 2 + mg( − A) 2( ∆L)
The mg term cancels, and the equation can be rearranged into the quadratic equation
( ∆L) 2 + 2 h( ∆L) − A 2 = 0 The positive solution is
∆L = h 2 + A 2 − h = (0.030 m) 2 + (0.100 m) 2 − 0.030 m = 0.0744 m Now that ∆L is known, we can find
ω=
g = ∆L
9.80 m / s 2 ω = 11.48 rad / s ⇒ f = = 1.83 Hz 0.0744 m 2π
14.78. Model: Model the bungee cord as a spring. The motion is damped SHM. Visualize:
Solve:
(a) For light damping, the oscillation period is
m 2π 2π ⇒ k = m = (75 kg) = 185 N / m T 4.0 s k 2
T = 2π
2
(b) The maximum speed is vmax = ωA = (2π/4.0 s)(11.0 m) = 17.3 m/s. (c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces, ∆L = mg/k = (75 kg)(9.80 m/s2)/(185 N/m) = 3.97 m. Jose’s lowest point is 11.0 m below this point, so the bungee cord is stretched by ∆ymax = ∆L + A = 14.97 m. Choose this lowest point as y = 0. Because Jose is instantaneously at rest at this point, his energy is entirely the elastic potential energy of the stretched bungee cord. Initially, his energy was entirely gravitational potential energy. Equating his initial energy to his energy at the lowest point,
U lowest point = U highest point ⇒ 12 k ( ∆ymax ) 2 = mgh h=
k ( ∆ymax ) 2 (185 N / m) (14.97 m) 2 = = 28.2 m 2 mg 2(75 kg)(9.80 m / s 2 )
Jose jumped 28.2 m above the lowest point. (d) The amplitude decreases due to damping as A(t) = Ae–bt/2m. At the time when the amplitude has decreased from 11.0 m to 2.0 m, 2.0 m 2m 2 2(75 kg) = e − bt / 2 m ⇒ t = − ln =− ( −1.705) = 42.6 s b 11 11.0 m 6.0 kg / s With a period of 4.0 s, the number of oscillations is Nosc = (42.6 s)/(4.0 s) = 10.7 oscillations.
14.79. Model: The vertical movement of the car is simple harmonic motion. Visualize:
The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the road provide a periodic external force to the car’s suspension system, and a resonance will occur when the “bump frequency” fext matches the car’s natural oscillation frequency f0. Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps are spaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period: ∆x 3.0 m T= = = 0.60 s v 5.0 m s The external frequency due to the bumps is thus fext = 1 T = 1.667 Hz . This matches the car’s natural frequency f0, which is the frequency the car oscillates up and down with if you push the car down and release it. This is enough information to deduce the spring constant of the car’s suspension:
1 k N 2 ⇒ k = m(2πfext ) = 131, 600 2π m m where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with Fnet = 0 N. The downward weight mtotalg of the car and passengers is balanced by the upward spring force k∆y of the suspension. Thus the compression ∆y of the suspension is m g ∆y = total k Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression ∆yi = 0.0894 m = 8.94 cm. When three additional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is ∆yf = 0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to “sag” by 11.17 cm – 8.94 cm = 2.23 cm. 1.667 Hz =
14.80. Solve: The solution of the equation d 2 x b dx k + + x=0 dt 2 m dt m is x (t ) = Ae − bt 2 m cos(ωt + φ 0 ). The first and second derivatives of x(t) are
dx Ab − bt 2 m =− cos(ωt + φ 0 ) − Aωe − bt 2 m sin(ωt + φ 0 ) e dt 2m ωAb d 2 x Ab 2 = − Aω 2 cos(ωt + φ 0 ) + sin(ωt + φ 0 )e − bt 2 m dt 2 4 m 2 m Substituting these expressions into the differential equation, the terms involving sin(ωt + φ0) cancel and we obtain the simplified result
−b 2 k − ω 2 + cos(ωt + φ 0 ) = 0 2 m 4m Because cos(ωt + φ 0 ) is not equal to zero in general,
−b 2 k −ω2 + = 0 ⇒ ω = 4m 2 m
k b2 − m 4m 2