An object of mass 0.2 ~kg executes simple harmonic motion along x axis with frequency of (\frac25\pi) 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 oscillation is \underline\hspace2cmcm .

Simple Harmonic Motion PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q85 with Solution

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Correct answer: 6

Mass: $m = 0.2\,\text{kg}$
Frequency: $f = \dfrac{25}{\pi}\,\text{Hz}$
At position $x = 0.04\,\text{m}$ $x = 0.04 m$ :
- Kinetic energy: $K = 0.5\,\text{J}$
- Potential energy: $U = 0.4\,\text{J}$

We need the amplitude $A$ .

\omega = 2\pi f = 2\pi \left(\frac{25}{\pi}\right) = 50\,\text{rad/s}

For SHM,

U = \frac{1}{2}m\omega^2 x^2

Check with given values:

U = \frac{1}{2}(0.2)(50)^2(0.04)^2

= 0.1 \cdot 2500 \cdot 0.0016

= 0.4\,\text{J}

This matches the given data.

Total mechanical energy in SHM is

E = K + U = 0.5 + 0.4 = 0.9\,\text{J}

Also,

E = \frac{1}{2}m\omega^2 A^2

So,

0.9 = \frac{1}{2}(0.2)(50)^2 A^2

0.9 = 0.1 \cdot 2500 \cdot A^2

0.9 = 250A^2

A^2 = \frac{0.9}{250} = 0.0036

A = \sqrt{0.0036} = 0.06\,\text{m}

0.06\,\text{m} = 6\,\text{cm}

The amplitude of oscillation is

\boxed{6\,\text{cm}}

The required integer answer is $\boxed{6}$ .