Question

A particle executes SHM with a time period \(T\). If its maximum acceleration is doubled keeping the amplitude constant, its new time period is}

• A. [option in figure]
• B. [option in figure]
• C. [option in figure]
• D. [option in figure]
• E. [option in figure]

Hint:

For SHM: \[ a_{\max}=\omega^2 A \] If amplitude is fixed, then \(a_{\max}\propto \omega^2\).

Answer 1

The Correct Option is D

In SHM, maximum acceleration is: \[ a_{\max}=\omega^2 A \] Since amplitude \(A\) is constant and \(a_{\max}\) is doubled: \[ \omega'^2 A = 2\omega^2 A \] \[ \omega'^2=2\omega^2 \] \[ \omega'=\sqrt{2}\,\omega \] Now time period is: \[ T=\frac{2\pi}{\omega} \] So new time period: \[ T'=\frac{2\pi}{\omega'}=\frac{2\pi}{\sqrt{2}\omega}=\frac{T}{\sqrt{2}} \]
Hence, the correct answer is the option corresponding to: \[ \boxed{\frac{T}{\sqrt{2}}} \]