JEE Physics 2024 Question with Solution

Given: Mass of particle is $m = 0.50 \, \text{kg}$ and restoring force is $F = -50x$, so comparing with $F = -kx$, we get $k = 50 \, \text{N m}^{-1}$.

Find: The value of $x$ if the time period is $\frac{x}{35} \, \text{s}$.

For simple harmonic motion,

$$\omega = \sqrt{\frac{k}{m}}$$

Substituting the given values,

$$\omega = \sqrt{\frac{50}{0.50}} = \sqrt{100} = 10 \, \text{rad/s}$$

The time period is

$$T = \frac{2\pi}{\omega}$$

So,

$$T = \frac{2\pi}{10}$$

Using $\pi = \frac{22}{7}$,

$$T = \frac{2 \times \frac{22}{7}}{10} = \frac{44}{70} = \frac{22}{35} \, \text{s}$$

Given,

$$T = \frac{x}{35} \, \text{s}$$

Equating,

$$\frac{x}{35} = \frac{22}{35}$$

Hence,

$$x = 22$$

Therefore, the value of $x$ is $22$ and the correct option is B.

Given: $m = 0.50 \, \text{kg}$ and $k = 50 \, \text{N m}^{-1}$ from $F = -kx$.

Find: The value of $x$ in $T = \frac{x}{35}$.

A direct formula for time period in simple harmonic motion is

$$T = 2\pi \sqrt{\frac{m}{k}}$$

Substitute the values,

$$T = 2\pi \sqrt{\frac{0.5}{50}} = 2\pi \sqrt{0.01} = 2\pi \times 0.1 = 0.2\pi \, \text{s}$$

Now use $T = \frac{x}{35}$,

$$0.2\pi = \frac{x}{35}$$

With $\pi = \frac{22}{7}$,

$$0.2 \times \frac{22}{7} = \frac{x}{35}$$

Thus,

$$x = 0.2 \times 22 \times 5 = 22$$

Therefore, the value of $x$ is $22$.