JEE Physics 2023 Question with Solution

Given: mass of the block is $5 \, \text{kg}$, amplitude is $1 \, \text{m}$, and time period is $3.14 \, \text{s}$.

Find: the maximum force exerted by the spring.

For simple harmonic motion,

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

Substituting $T = 3.14 \, \text{s}$,

$$\omega = \frac{2 \times \frac{22}{7}}{3.14} = 2 \, \text{rad/s}$$

The maximum acceleration is,

$$a_{\text{max}} = \omega^2 A$$

Substituting $\omega = 2 \, \text{rad/s}$ and $A = 1 \, \text{m}$,

$$a_{\text{max}} = 2^2 \times 1 = 4 \, \text{m/s}^2$$

Now the maximum force is,

$$F_{\text{max}} = m a_{\text{max}}$$

Substituting $m = 5 \, \text{kg}$ and $a_{\text{max}} = 4 \, \text{m/s}^2$,

$$F_{\text{max}} = 5 \times 4 = 20 \, \text{N}$$

Therefore, the maximum force exerted by the spring is $20 \, \text{N}$.

Given: $m = 5 \, \text{kg}$, $A = 1 \, \text{m}$, and $T = 3.14 \, \text{s}$.

Find: maximum restoring force.

In simple harmonic motion, restoring force is maximum at the extreme position. Since

$$a_{\text{max}} = \omega^2 A$$

and

$$F_{\text{max}} = m a_{\text{max}}$$

we get

$$F_{\text{max}} = m\omega^2 A$$

Now,

$$\omega = \frac{2\pi}{T} = \frac{2\pi}{3.14} \approx 2 \, \text{rad/s}$$

Hence,

$$F_{\text{max}} = 5 \times (2)^2 \times 1 = 20 \, \text{N}$$

So, the required numerical value is 20.