The amplitude of a particle executing SHM is $3 \mathrm{~cm}$. The displacement at which its kinetic energy will be $25 \%$ more than the potential energy is: __________ $\mathrm{cm}$
Answer (integer)
2
Solution
$A=3 \mathrm{~cm}$
<br/><br/>$$
\begin{aligned}
& K=1.25 U \\\\
& \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\
& \Rightarrow \frac{9}{5} K=K_{\max } \\\\
& \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\
& \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\
& \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\
& \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\
& \Rightarrow x=\frac{2 A}{3} \\\\
& \Rightarrow x=2 \mathrm{~cm}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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