JEE Physics 2023 Question with Solution

Given: The initial time period is $1 \, \text{s}$. After increasing the mass by $3 \, \text{kg}$, the new time period becomes $2 \, \text{s}$.

Find: The value of mass $m$.

For a mass-spring system, the time period is

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

Initially,

$$1 = 2\pi\sqrt{\frac{m}{k}}$$

Squaring both sides,

$$\frac{m}{k} = \frac{1}{4\pi^{2}}$$

After increasing the mass by $3 \, \text{kg}$, the period becomes $2 \, \text{s}$:

$$2 = 2\pi\sqrt{\frac{m+3}{k}}$$

Squaring both sides,

$$\frac{m+3}{k} = \frac{4}{4\pi^{2}} = \frac{1}{\pi^{2}}$$

Now divide the second equation by the first:

$$\frac{m+3}{m} = 4$$

So,

$$m+3 = 4m$$ $$3m = 3$$ $$m = 1 \, \text{kg}$$

Therefore, the value of mass is $1$.

Given: $T=1 \, \text{s}$ initially and $T'=2 \, \text{s}$ after adding $3 \, \text{kg}$.

Find: The value of $m$.

Using $T \propto \sqrt{m}$ for the same spring constant,

$$\frac{T'}{T} = \sqrt{\frac{m+3}{m}}$$

Substitute the values:

$$\frac{2}{1} = \sqrt{\frac{m+3}{m}}$$

Squaring,

$$4 = \frac{m+3}{m}$$ $$m+3 = 4m$$ $$m = 1$$

Therefore, the required mass is $1$.