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Q.
For a SHM, if the maximum potential energy become double, choose the correct statements.
Oscillations
Solution:
For a particle executing SHM, its kinetic energy,
$KE =\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)$
$\Rightarrow ( KE )_{\max }=\frac{1}{2} m \omega^{2} A^{2}$
Potential energy, $PE =( PE )_{\max } =\frac{1}{2} m \omega^{2} A^{2}$
Total mechanical energy $TE = KE + PE$
$ =\frac{1}{2} m \omega^{2} A^{2} $
$\Rightarrow TE =( KE )_{\max } =( PE )_{\max }$
So, if maximum potential energy becomes double, then both total energy and kinetic energy will also become double.
Thus, the statements given in both options (a) and (b) are correct.