Discussion
A particle executes SHM with a time period T. If its maximum acceleration is doubled keeping the amplitude constant, its new time period is
Answer: D
The new time period is \(\frac{T}{\sqrt{2}}\).
In SHM the maximum acceleration is \(a_{max} = \omega^{2}A\). With A fixed, doubling \(a_{max}\) doubles \(\omega^{2}\).
\[ \begin{aligned} \omega'^{2} &= 2\omega^{2} \Rightarrow \omega' = \sqrt{2}\,\omega \\ T' &= \frac{2\pi}{\omega'} = \frac{T}{\sqrt{2}} \end{aligned} \]
\(\frac{T}{2}\) would need \(\omega\) to double, which means \(a_{max}\) four times larger. \(\sqrt{2}T\) and 2T go the wrong way: a larger restoring acceleration makes the motion faster, not slower.
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