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The maximum acceleration of a particle in SHM is made four times keeping the maximum speed to be constant. It is possible when

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amplitude of oscillation is made one-fourth while frequency is made four times

amplitude of oscillation is doubled while frequency remains constant

amplitude of oscillation remains constant while frequency is doubled

amplitude of oscillation is doubled while frequency is halved

answer is C.

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Detailed Solution

$\begin{array}{l} a_{max} = A ω^{2} . . . . . . (1) \\ v_{max} = A ω . . . . . . . . (2) \\ \Rightarrow a_{max} = v_{max} ω \end{array}$

Since, $v_{\max}$ is constant, to make $a_{\max}$ four times, $ω$ should be made four times.
From (1), A should be made one-fourth.

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