At a given point of time the value of… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Waves & Oscillations

2023integerhard

At a given point of time the value of displacement of a simple harmonic oscillator is given as $y=A\mathrm{cos}(30^{\circ}).$ If amplitude is $40\mathrm{cm}$ and kinetic energy at that time is $200J,$ the value of force constant $1.0\times {10}^{x}N{m}^{–1}$ . The value of $x$ is _____.

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Held on 13 Apr 2023 · Verified 6 Jul 2026.

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Solution

The formula to calculate the kinetic energy $(K)$ of a particle executing SHM is given by

$K=\frac{1}{2}k({A}^{2}-{y}^{2})...(1)$

Substitute the values of the known parameters into equation (1) and solve to calculate the value of the unknown parameter.

$\begin{matrix}K & = & \frac{1}{2}\times k\times ({A}^{2}-{A}^{2}{\mathrm{cos}}^{2}30^{\circ}) \\ & = & \frac{1}{2}\times k\times {A}^{2}{\mathrm{sin}}^{2}30^{\circ} \\ & \Rightarrow & 200=\frac{1}{2}\times 1.0\times {10}^{x}\times {(0.4)}^{2}\times {(\frac{1}{2})}^{2} \\ & = & \frac{2\times {10}^{x}}{100} \\ & \Rightarrow & {10}^{2}={10}^{x-2} \\ & \Rightarrow & x-2=2 \\ & \Rightarrow & x=4\end{matrix}$

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