JEE Physics 2023 Question with Solution

Given: Mass of the block is $m = 2 \, \text{kg}$. Two identical springs are attached, each with spring constant $k_1 = k_2 = 20 \, \text{N/m}$.

Find: The value of $x$ in $T = \pi/\sqrt{x}$.

For this arrangement, the equivalent spring constant is

$$k_{\text{eq}} = k_1 + k_2 = 20 + 20 = 40 \, \text{N/m}$$

The angular frequency of the block is

$$\omega = \sqrt{\frac{k_{\text{eq}}}{m}} = \sqrt{\frac{40}{2}} = \sqrt{20} \, \text{rad/s}$$

Hence, the time period is

$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{20}} = \frac{\pi}{\sqrt{5}} \, \text{s}$$

Comparing with $T = \pi/\sqrt{x}$, we get

$$x = 5$$

Therefore, the value of $x$ is $5$.

Given: Two identical springs of constant $20 \, \text{N/m}$ each are attached to a block of mass $2 \, \text{kg}$.

Find: The value of $x$.

When the block is displaced, both springs provide restoring force together, so the effective spring constant becomes

$$k_{\text{eq}} = 20 + 20 = 40 \, \text{N/m}$$

Now use

$$T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}}$$

Substituting,

$$T = 2\pi \sqrt{\frac{2}{40}} = 2\pi \sqrt{\frac{1}{20}} = \frac{\pi}{\sqrt{5}}$$

So, comparing with $\pi/\sqrt{x}$, the correct value is $x = 5$. Therefore, the answer is $5$.