Question

An object is executing simple harmonic motion with an angular frequency \(\omega\). If the maximum velocity is \(v_{\max}\), then the maximum acceleration of the object is

• A. \(\omega^2v_{\max}\)
• B. \(\omega v_{\max}\)
• C. \(\omega\sqrt{v_{\max}}\)
• D. \(3\omega v_{\max}\)

Hint:

Remember the standard SHM relations: \[ v_{\max}=\omega A \] and \[ a_{\max}=\omega^2A \] Using these, eliminate amplitude whenever required.

Answer 1

The Correct Option is B

Step 1: Recall the expressions for maximum velocity and acceleration in SHM.
For SHM, \[ v_{\max}=\omega A \] and maximum acceleration is \[ a_{\max}=\omega^2A \]

Step 2: Eliminate amplitude \(A\).
From \[ v_{\max}=\omega A, \] we get \[ A=\frac{v_{\max}}{\omega} \] Substitute into \[ a_{\max}=\omega^2A \] \[ a_{\max}=\omega^2\left(\frac{v_{\max}}{\omega}\right) \] \[ a_{\max}=\omega v_{\max} \]

Step 3: Final conclusion.
Hence, the maximum acceleration is \[ \boxed{\omega v_{\max}} \]