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multiple attempts not allowed. this test can only be taken once.
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this final exam is cumulative and covers material from the entire course.
from the list of choices, select the one best answer.
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question 53
the displacement of a mass attached to the end of a vibrating, ideal spring is given by:
x(t) = (0.18 m)cos(π/3 t)
what is the magnitude of the maximum velocity of the mass?
0.36 m/s
0.72 m/s
0.19 m/s
1.0 m/s
Step1: Recall velocity from displacement
The velocity \( v(t) \) is the derivative of the displacement \( x(t) \) with respect to time \( t \). Given \( x(t) = A\cos(\omega t) \), where \( A = 0.18\space m \) and \( \omega=\frac{\pi}{3}\space rad/s \), the derivative is \( v(t)=-A\omega\sin(\omega t) \).
Step2: Find maximum velocity magnitude
The maximum value of \( |\sin(\omega t)| \) is 1. So the maximum velocity magnitude \( v_{max}=A\omega \). Substitute \( A = 0.18\space m \) and \( \omega=\frac{\pi}{3}\space rad/s \): \( v_{max}=0.18\times\frac{\pi}{3} \). Calculate \( 0.18\times\frac{\pi}{3}=0.06\pi\approx0.188\approx0.19\space m/s \)? Wait, no, wait: Wait, \( 0.18\times\frac{\pi}{3}=0.06\pi\approx0.188 \), but wait, maybe I miscalculated. Wait, \( \frac{\pi}{3}\approx1.047 \), so \( 0.18\times1.047\approx0.188\approx0.19 \)? Wait, but let's recalculate: \( 0.18\times\frac{\pi}{3}=0.06\pi\approx0.1884955\approx0.19\space m/s \). Wait, but let's check the options. Wait, maybe I made a mistake. Wait, no, wait: Wait, the formula for maximum velocity in simple harmonic motion is \( v_{max}=A\omega \). So \( A = 0.18 \), \( \omega=\frac{\pi}{3} \). So \( v_{max}=0.18\times\frac{\pi}{3}=0.06\pi\approx0.188\approx0.19\space m/s \). So the correct option is 0.19 m/s. Wait, but let's check again. Wait, \( 0.18\times\frac{\pi}{3}=0.06\pi\approx0.188 \), which is approximately 0.19 m/s. So the answer should be 0.19 m/s.
Wait, no, wait, maybe I messed up the derivative. Wait, \( x(t)=A\cos(\omega t) \), so \( v(t)=\frac{dx}{dt}=-A\omega\sin(\omega t) \). The maximum value of \( |\sin(\omega t)| \) is 1, so \( v_{max}=A\omega \). So \( A = 0.18 \), \( \omega=\frac{\pi}{3} \). So \( v_{max}=0.18\times\frac{\pi}{3}=0.06\pi\approx0.188\approx0.19\space m/s \). So the correct option is 0.19 m/s.
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0.19 m/s (corresponding to the option "0.19 m/s")