An object of mass 0.2 ~kg executes simple harmonic motion along x axis with frequency of ((25)/(π)) Hz . At the position x=0.04 ~m the object has kinetic energy 0.5 ~J and potential energy 0.4 ~J . The amplitude of

Correct answer: 0 aligned & Total energy = K.E. + P.E. & at x=0.04 m, T.E. =0.5+0.4=0.9 J & aligned T.E & =1 m ω^2 A^2=0.9 & =(1)/(2) × 0.2(2 π × (25)/(π))^2 × A^2=0.9 A & =0.06 m A & =6 cm aligned aligned

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Oscillations

JEE Main 2024 · 8 April · Evening Shift · Physics · Oscillations

An object of mass 0.2 ~kg executes simple harmonic motion along x...

An object of mass

0.2 \mathrm{~kg}

executes simple harmonic motion along

\mathrm{x}

axis with frequency of \left(\frac{25}{\pi}\right) \mathrm{Hz}

. At the position

x=0.04 \mathrm{~m}

the object has kinetic energy

0.5 \mathrm{~J} and potential energy

0.4 \mathrm{~J}

. The amplitude of oscillation is _____

\mathrm{cm}

Correct answer

Step-by-step solution

\begin{aligned} & \text{ Total energy }=\text{ K.E. }+ \text{ P.E. } \\ & \text{ at } x=0.04 \mathrm{ m}, \text{ T.E. }=0.5+0.4=0.9 \mathrm{ J} \\ & \begin{aligned} \text{ T.E } & =1 \mathrm{ m} \omega^2 \mathrm{ A}^2=0.9 \\ & =\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^2 \times \mathrm{A}^2=0.9 \\ \Rightarrow \mathrm{A} & =0.06 \mathrm{ m} \\ \mathrm{ A} & =6 \mathrm{ cm}\end{aligned}

\end{aligned}$

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