The maximum acceleration of a particle in SHM | vedclass

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Question

(C) The maximum acceleration is given by $a_{\max} = A\omega^2$ and the maximum speed is given by $v_{\max} = A\omega$.Let the initial amplitude be $A

Vedclass · Accepted Answer

frequency is doubled while amplitude is halved

Vedclass · Answer

(C) The maximum acceleration is given by $a_{\max} = A\omega^2$ and the maximum speed is given by $v_{\max} = A\omega$.Let the initial amplitude be $A_1$ and angular frequency be $\omega_1$. Then $v_{\max} = A_1\omega_1$ and $a_{\max} = A_1\omega_1^2$.If the new amplitude is $A_2$ and new angular frequency is $\omega_2$,we are given that $v_{\max}$ remains constant,so $A_1\omega_1 = A_2\omega_2$.We are also given that the new maximum acceleration is twice the original,so $A_2\omega_2^2 = 2(A_1\omega_1^2)$.Dividing the second equation by the first: $\frac{A_2\omega_2^2}{A_1\omega_1} = \frac{2A_1\omega_1^2}{A_1\omega_1} \implies \frac{\omega_2}{\omega_1} = 2$,which means $\omega_2 = 2\omega_1$ (frequency is doubled).Substituting $\omega_2 = 2\omega_1$ into $A_1\omega_1 = A_2\omega_2$,we get $A_1\omega_1 = A_2(2\omega_1) \implies A_2 = \frac{A_1}{2}$ (amplitude is halved).Therefore,the frequency is doubled and the amplitude is halved.

The maximum acceleration of a particle in $SHM$ is made two times keeping the maximum speed constant. This is possible when:

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