JEE Physics 2025 Question with Solution

Given: $x_1 = \sqrt{7} \sin 5t \, \text{cm}$ $x_2 = 2\sqrt{7} \sin\left(5t + \frac{\pi}{3}\right) \, \text{cm}$ Both SHMs have the same angular frequency $\omega = 5 \, \text{rad/s}$.

Find: The value of $x$ if maximum acceleration is x \times 10^{-2} \, \text{m/s^2}.

For superposition of two SHMs of the same frequency, the resultant motion is also SHM. Its amplitude is

$$A_R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\Delta\phi}$$

Here,

$$A_1 = \sqrt{7} \;, \quad A_2 = 2\sqrt{7} \;, \quad \Delta\phi = \frac{\pi}{3}$$

So,

$$A_R^2 = (\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7})\cos\left(\frac{\pi}{3}\right)$$ $$A_R^2 = 7 + 28 + 28\left(\frac{1}{2}\right)$$ $$A_R^2 = 7 + 28 + 14 = 49$$ $$A_R = 7 \, \text{cm}$$

Convert amplitude into SI unit:

$$A_R = 7 \, \text{cm} = 0.07 \, \text{m}$$

Maximum acceleration in SHM is

$$a_{\max} = \omega^2 A_R$$

Hence,

$$a_{\max} = (5)^2 \times 0.07$$ $$a_{\max} = 25 \times 0.07 = 1.75 \, \text{m/s}^2$$

Given,

$$x \times 10^{-2} = 1.75$$

Therefore,

$$x = 175$$

So, the correct option is A.

Given: The two SHM amplitudes are represented by phasors of magnitudes $\sqrt{7}$ and $2\sqrt{7}$ with phase angle $60^\circ$ between them.

Find: Resultant maximum acceleration.

From phasor addition, the resultant amplitude is

$$A_R = \sqrt{(\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7})\cos 60^\circ}$$ $$A_R = \sqrt{7 + 28 + 14} = \sqrt{49} = 7 \, \text{cm}$$

Thus the resultant SHM can be written as

$$X_R = 7 \sin(5t + \phi)$$

for some phase $\phi$. Its acceleration is

$$a_R = -\omega^2 X_R$$

So the maximum magnitude is

$$a_{\max} = \omega^2 A_R = 25 \times 7 = 175 \, \text{cm/s}^2$$

Now,

$$175 \, \text{cm/s}^2 = 175 \times 10^{-2} \, \text{m/s}^2$$

Therefore, the required value is $175$, so the correct option is A.