JEE Main 2023 — Physics Waves & Oscillations

2023integereasy

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: ______ $\mathrm{cm}$ .

Official previous-year question

Held on 1 Feb 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Let $x$ be the displacement at which its kinetic energy will be $25%$ more than the potential energy.

Given here, $A=3\mathrm{cm}$ and $KE=PE+\frac{PE}{4}$

So, $KE=\frac{5}{4}PE$

$\Rightarrow \frac{1}{2}m{\omega }^{2}({A}^{2}-{x}^{2})=\frac{5}{4}\times \frac{1}{2}m{\omega }^{2}{x}^{2}$

$\Rightarrow {A}^{2}-{x}^{2}=\frac{5}{4}{x}^{2}$

$\Rightarrow \frac{9{x}^{2}}{4}={A}^{2}$

$\Rightarrow x=\frac{2}{3}A$

$\therefore x=\frac{2}{3}\times 3\mathrm{cm}$

$\Rightarrow x=2\mathrm{cm}$

Your note

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JEE Main 2023 — Physics Waves & Oscillations

2023integereasy

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: ______ $\mathrm{cm}$ .

Official previous-year question

Held on 1 Feb 2023 · Verified 6 Jul 2026.

Did you get this right?

Solution

Let $x$ be the displacement at which its kinetic energy will be $25%$ more than the potential energy.

Given here, $A=3\mathrm{cm}$ and $KE=PE+\frac{PE}{4}$

So, $KE=\frac{5}{4}PE$

$\Rightarrow \frac{1}{2}m{\omega }^{2}({A}^{2}-{x}^{2})=\frac{5}{4}\times \frac{1}{2}m{\omega }^{2}{x}^{2}$

$\Rightarrow {A}^{2}-{x}^{2}=\frac{5}{4}{x}^{2}$

$\Rightarrow \frac{9{x}^{2}}{4}={A}^{2}$

$\Rightarrow x=\frac{2}{3}A$

$\therefore x=\frac{2}{3}\times 3\mathrm{cm}$

$\Rightarrow x=2\mathrm{cm}$

Your note

JEE Main Waves & Oscillations in other years

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The equation of a plane progressive wave is given by $y = 5\cos\pi\left(200t - \dfrac{x}{150}\right)$ where $x$ and $y$ are in cm and $t$ is in second. The velocity of the wave is _______ m/s.

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