JEE Physics 2023 Question with Solution

Given: Mass of the particle is $m = 0.25 \, \text{kg}$, force law is $F = -25x$, and maximum speed is $v_{\text{max}} = 4 \, \text{m/s}$.

Find: The amplitude of the SHM in centimetres.

For simple harmonic motion,

$$F = -kx$$

and

$$a = -\omega^2 x$$

So,

$$\omega^2 = \frac{k}{m}$$

From $F = -25x$, we get

$$k = 25 \, \text{N/m}$$

Now substitute $m = 0.25 \, \text{kg}$:

$$\omega^2 = \frac{25}{0.25} = 100$$

Hence,

$$\omega = 10 \, \text{rad/s}$$

The maximum speed in SHM is

$$v_{\text{max}} = A\omega$$

Therefore,

$$A = \frac{v_{\text{max}}}{\omega} = \frac{4}{10} = 0.4 \, \text{m}$$

Converting to centimetres,

$$0.4 \, \text{m} = 40 \, \text{cm}$$

Therefore, the amplitude of the motion is $40 \, \text{cm}$.

Given: $k = 25 \, \text{N/m}$, $m = 0.25 \, \text{kg}$, and $v_{\text{max}} = 4 \, \text{m/s}$.

Find: Amplitude $A$.

In SHM,

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{25}{0.25}} = \sqrt{100} = 10 \, \text{rad/s}$$

Then use

$$A = \frac{v_{\text{max}}}{\omega} = \frac{4}{10} = 0.4 \, \text{m} = 40 \, \text{cm}$$

This works because the particle has its greatest speed at the mean position, where the SHM relation $v_{\text{max}} = A\omega$ applies directly.

Therefore, the required answer is $40$.