A particle performing SHM with energy of vibration 90 J

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Kindly find the answer of your asked query.

$$\begin{gathered} Energy of the particle executing SHM is given by: \\ E=\frac{1}{2}K{A}^2 \\ Where A is the amplitude \\ A=6 cm and new amplitude A\text{'}=4 cm \\ New energy E\text{'} is gievn by: \\ \frac{E\text{'}}{E}=\frac{\frac{1}{2}KA{\text{'}}^2}{\frac{1}{2}KA{\text{'}}^2}=\frac{36}{16} \\ E\text{'}=\frac{16}{36}\times 90=40 Joule \end{gathered}$$

Option 1. 40 J

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