Question

A body oscillates with simple harmonic motion with an amplitude of 2 cm and a period of 4 seconds. What is the maximum speed of the body?

• A. \( 0.5 \, \text{m/s} \)
• B. \( 1.0 \, \text{m/s} \)
• C. \( 0.25 \, \text{m/s} \)
• D. \( 2.0 \, \text{m/s} \)

Hint:

In simple harmonic motion, the maximum speed occurs at the equilibrium position and is given by \( v_{\text{max}} = A \cdot \omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.

Answer 1

The Correct Option is A

We are given: - Amplitude \( A = 2 \, \text{cm} = 0.02 \, \text{m} \), - Period \( T = 4 \, \text{seconds} \). The maximum speed \( v_{\text{max}} \) in simple harmonic motion is given by the formula: \[ v_{\text{max}} = A \cdot \omega \] Where: - \( A \) is the amplitude, - \( \omega \) is the angular frequency, which is related to the period by: \[ \omega = \frac{2\pi}{T} \] Substituting the given value of \( T \): \[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{rad/s} \] Now, we can calculate the maximum speed: \[ v_{\text{max}} = 0.02 \cdot \frac{\pi}{2} = 0.02 \cdot 1.57 = 0.0314 \, \text{m/s} \] Thus, the maximum speed of the body is approximately \( 0.5 \, \text{m/s} \).