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QJEEM23Phbcdc: JEE Main 2023 Physics - Class 11 JEEM23Phbcdc Solution: The amplitude of a particle executing SHM is . The displacement …

The amplitude of a particle executing SHM is 3 cm3 \mathrm{~cm} . The displacement at which its kinetic energy will be 25%25 \% more than the potential energy is: __________ cm\mathrm{cm}

Step-by-Step Verified Solution

A=3 cmA=3 \mathrm{~cm}

amp;K=1.25Uamp;⇒K+K1.25=Kmax⁡amp;⇒95K=Kmax⁡amp;⇒9512mv2=12mvmax⁡2amp;⇒95[ωA2−x2]2=ω2A2amp;⇒9(A2−x2)=5A2amp;⇒x2=4A29amp;⇒x=2A3amp;⇒x=2 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}

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