JEE Physics 2023 Question with Solution

Given: The amplitude is $A = 3 \, \text{cm}$.

Find: The displacement $x$ at which kinetic energy is $25\%$ more than potential energy.

From the condition,

$$KE = PE + \frac{PE}{4}$$

So,

$$KE = \frac{5}{4} PE$$

For SHM,

$$KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$$

and

$$PE = \frac{1}{2} m \omega^2 x^2$$

Substituting in the ratio,

$$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{5}{4} \times \frac{1}{2} m \omega^2 x^2$$

Hence,

$$A^2 - x^2 = \frac{5}{4} x^2$$ $$A^2 = \frac{9}{4} x^2$$ $$x = \frac{2}{3} A$$

Now substitute $A = 3 \, \text{cm}$:

$$x = \frac{2}{3} \times 3 \, \text{cm} = 2 \, \text{cm}$$

Therefore, the displacement is $2 \, \text{cm}$ and the correct option is B.

Given: A particle executes SHM with amplitude $3 \, \text{cm}$.

Find: The displacement where kinetic energy exceeds potential energy by $25\%$.

Potential energy at displacement $x$ is proportional to $x^2$, while kinetic energy is proportional to $(A^2 - x^2)$.

Using the condition,

$$KE = 1.25 \, PE = \frac{5}{4} PE$$

Now write both energies:

$$\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{5}{4} \left( \frac{1}{2} m \omega^2 x^2 \right)$$

Cancel the common factor $\frac{1}{2} m \omega^2$:

$$A^2 - x^2 = \frac{5}{4} x^2$$

Bring terms together:

$$A^2 = x^2 + \frac{5}{4} x^2 = \frac{9}{4} x^2$$ $$x^2 = \frac{4}{9} A^2$$ $$x = \frac{2}{3} A$$

Since $A = 3 \, \text{cm}$,

$$x = \frac{2}{3} \times 3 = 2 \, \text{cm}$$

Therefore, the required displacement is $2 \, \text{cm}$.