patel (rpp467) – Quest #9 SHM (part 3/3) – ha – (51122013) This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
x(t) 2 1
001 10.0 points Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. To make sure that their products meet this specification, manufacturers test them using a shaking table that can vibrate a device at various specified frequencies and amplitudes. If a device is given a vibration of amplitude 6.6 cm, what should be its frequency in order to test for compliance with the 10 g military specification? The acceleration of gravity is 9.81 m/s2 .
1. 2.0 rad/s
Correct answer: 6.13596 Hz.
5. 1.6 rad/s correct
1 2 3 4 5 6 7 8 9 10 11 12 t (sec) What is the angular frequency ω?
2. 6.2 rad/s 3. 4.0 rad/s 4. 3.1 rad/s
Explanation: From the graph, T = 4 s.
Explanation:
Let :
1
2
amax = 10 g = 98.1 m/s and A = 6.6 cm = 0.066 m .
The angular frequency is ω = 2 πf and the maximum accleration of the oscillator is
2π 2π = T 4 sec π = 1/sec = 1.5708 1/s . 2
ω=
003 (part 1 of 4) 10.0 points A 0.7 kg block attached to a spring of force constant 9.5 N/m oscillates with an amplitude of 20 cm. Find the maximum speed of the block. Correct answer: 0.736788 m/s. Explanation:
amax = A ω 2 = 4 π 2A f 2 r 1 amax f= 2π A r 98.1 m/s2 1 = 2π 0.066 m = 6.13596 Hz .
Let : A = 20 cm = 0.2 m , k = 9.5 N/m , and m = 0.7 kg . vmax = A ω r =A
k m s
= (0.2 m) 002 10.0 points An oscillator is described by
9.5 N/m 0.7 kg
= 0.736788 m/s .
patel (rpp467) – Quest #9 SHM (part 3/3) – ha – (51122013) 004 (part 2 of 4) 10.0 points Find the speed of the block when it is 10 cm from the equilibrium position.
Explanation: A . 2
Since x = A cos (ω t)
Thus
Explanation: When x = 0, 0 = A cos(ωt) π . 2 π π For ∆t = time to go from ω t = to ω t = , 3 2 π π π so ω ∆t = − = 2 3 6 π ∆t = 6 ωr π m = 6 k s π 0.7 kg = 6 9.5 N/m ω t = cos−1 0 =
Correct answer: 0.638077 m/s.
Let : x = 10 cm =
A = A cos(ω t) 2 1 = cos ωt and 2 π −1 1 = . ω t = cos 2 3 v = vmax sin ωt = (0.736788 m/s) sin
π 3
= 0.638077 m/s . 005 (part 3 of 4) 10.0 points Find its acceleration at 10 cm from the equilibrium position. Correct answer: 1.35714 m/s2 .
= 0.14213 s . 007 10.0 points A 56.3 g object is attached to a horizontal spring with a spring constant of 11.9 N/m and released from rest with an amplitude of 24.4 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless? Correct answer: 3.07213 m/s. Explanation:
Explanation: a = A ω 2 cos(ω t) k =A cos(ω t) m π 9.5 N/m cos = (0.2 m) 0.7 kg 3 = 1.35714 m/s2 .
Let : m = 56.3 g = 0.0563 kg , k = 11.9 N/m , and A = 24.4 cm = 0.244 m . The speed is r k 2 v= (A − x2 ) m s =
006 (part 4 of 4) 10.0 points Find the time it takes the block to move from x = 0 to x = 10 cm. Correct answer: 0.14213 s.
2
11.9 N/m [(0.244 m)2 − (0.122 m)2 ] 0.0563 kg
= 3.07213 m/s . 008
10.0 points
patel (rpp467) – Quest #9 SHM (part 3/3) – ha – (51122013) A 4 kg object oscillates on a spring with an amplitude of 84.7 cm with a maximum acceleration of 8.8 m/s2 . Find the total energy.
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010 (part 2 of 2) 10.0 points Find the amplitude of the motion. Correct answer: 24.1448 cm.
Correct answer: 14.9072 J.
Explanation:
Explanation: Given : E = 2.85 J . Let : m = 4 kg , A = 84.7 cm = 0.847 m , amax = 8.8 m/s2 .
and
At the maximum amplitude, the object is momentarily at rest, so 1 1 m v 2 + k A2 2 2 1 = k A2 , 2
E =K +U =
The maximum speed is vmax = A ω and the maximum acceleration is amax = A ω 2 A amax = A2 ω 2 . The total energy is 1 1 1 2 Etot = m vmax = m (A ω)2 = m A amax 2 2 2 1 = (4 kg) (0.847 m) (8.8 m/s2 ) 2 = 14.9072 J .
since v = 0, so the amplitude is r 2E A= s k 100 cm 2 (2.85 J) = (97.7746 N/m) m = 24.1448 cm .
009 (part 1 of 2) 10.0 points A(n) 204 g object is attached to a spring and executes simple harmonic motion with a period of 0.287 s. The total energy of the system is 2.85 J. Find the force constant of the spring.
011 (part 1 of 2) 10.0 points A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the ceiling almost to the floor and that its period is 16.8 s. How tall is the tower? The acceleration of gravity is 9.8 m/s2 .
Correct answer: 97.7746 N/m.
Correct answer: 70.0624 m.
Explanation:
Explanation:
Given : m = 204 g = 0.204 kg T = 0.287 s . The period is r m T = 2π k 2 4π m k= T2 4 π 2 (0.204 kg) = (0.287 s)2 = 97.7746 N/m .
and
Let :
T = 16.8 s .
The height of the tower is approximately the same as the length of the pendulum. The period of the pendulum is s L T = 2π g 9.8 m/s2 (16.8 s)2 g T2 = L= 4 π2 4 π2 = 70.0624 m .
patel (rpp467) – Quest #9 SHM (part 3/3) – ha – (51122013)
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The period of a physical pendulum is 012 (part 2 of 2) 10.0 points If this pendulum is taken to the Moon, where the free-fall acceleration is 1.67 m/s2 , what is its period there?
T = 2π
Explanation: d2 −
aM = 1.67 m/s2 .
The period of the pendulum on the moon is s s L 70.0624 m TMoon = 2 π = 2π aM 1.67 m/s2
I mgd
1 T2 mg d = m R 2 + m d2 2 4π 2
Correct answer: 40.6972 s.
Let :
s
g T2 R2 d + = 0. 4 π2 2
Applying the quadratic formula, g T2 d=
4 π2
±
s
= 40.6972 s .
g2 T 4 − 2 R2 4 16 π . 2
Since 013 (part 1 of 3) 10.0 points A uniform disk of radius 1 m and 1.1 kg mass has a small hole a distance d from the disk’s center that can serve as a pivot point.
d
1m
1.1 kg
What should be the distance d so that the period of this physical pendulum is 9.2 s? The acceleration due to gravity is 9.81 m/s2 .
g T2 (9.81 m/s2 ) (9.2 s)2 = 4 π2 4 π2 = 21.0322 m/s2 and
g2 T 4 (9.81 m/s2 )2 (9.2 s)4 2 − 2 R = 16 π 4 16 π 4 − 2 (1 m)2 = 440.354 m2 , then
21.0322 m/s2 − d= 2 = 0.0238 m .
√
440.354 m2
Correct answer: 0.0238 m. Explanation: Let : R = 1 m , m = 1.1 kg , T = 9.2 s , and g = 9.81 m/s2 .
014 (part 2 of 3) 10.0 points What should be the distance d so that this physical pendulum will have the shortest possible period? Correct answer: 0.707107 m. Explanation: The period is
Using the parallel-axis theorem, 1 I = Icm + m d = m R2 + m d2 . 2 2
T = 2π
s
R2 −1 d d + 2g g
patel (rpp467) – Quest #9 SHM (part 3/3) – ha – (51122013) and its derivative is −1/2 dT 1 R2 d = 2π + dd 2 2gd g R2 −2 1 × − . d + 2g g For minimum period, 0=−
dT = 0, so dd
R2 −2 1 d + 2g g
R2 d = 2 1m R d = √ = √ = 0.707107 m . 2 2 2
015 (part 3 of 3) 10.0 points What will be the period at this distance? Correct answer: 2.38563 s. Explanation: v u u1 2 u R + d2 t2 T = 2π gd v u u1 u (1 m)2 + (0.0238 m)2 t2 = 2π (9.81 m/s2 ) (0.0238 m) = 2.38563 s .
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