JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 27)
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
2
Explanation
$A=3 \mathrm{~cm}$
$$ \begin{aligned} & K=1.25 U \\\\ & \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\ & \Rightarrow \frac{9}{5} K=K_{\max } \\\\ & \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\ & \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\ & \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\ & \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\ & \Rightarrow x=\frac{2 A}{3} \\\\ & \Rightarrow x=2 \mathrm{~cm} \end{aligned} $$
$$ \begin{aligned} & K=1.25 U \\\\ & \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\ & \Rightarrow \frac{9}{5} K=K_{\max } \\\\ & \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\ & \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} \\\\ & \Rightarrow 9\left(A^{2}-x^{2}\right)=5 A^{2} \\\\ & \Rightarrow x^{2}=\frac{4 A^{2}}{9} \\\\ & \Rightarrow x=\frac{2 A}{3} \\\\ & \Rightarrow x=2 \mathrm{~cm} \end{aligned} $$
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