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question 12
problem reference 10.2
a simple harmonic oscillator vibrates back and forth and its displacement as a function of time is given by
x(t) = 0.500 m cos( (π/3)t )
what is the magnitude of the maximum velocity?
9.18 m/s
4.67 m/s
0.214 m/s
0.524 m/s

Explanation:

Step1: Recall SHM velocity formula

Velocity \( v(t) = \frac{dx(t)}{dt} = -A\omega \sin(\omega t) \)

Step2: Identify \( A \) and \( \omega \)

From \( x(t)=0.500\cos(\frac{\pi}{3}t) \), \( A=0.500\,\text{m} \), \( \omega=\frac{\pi}{3}\,\text{rad/s} \)

Step3: Calculate max velocity

Max \( |v| = A\omega = 0.500 \times \frac{\pi}{3} \approx 0.524\,\text{m/s} \)

Answer:

0.524 m/s