JEE Physics 2024 Question with Solution

Given: A particle is in SHM with amplitude $A$. At displacement $x = \frac{2A}{3}$, its speed becomes three times the original speed.

Find: If the new amplitude is $\frac{nA}{3}$, find $n$.

For SHM, the speed at displacement $x$ is

$$v = \omega \sqrt{A^2 - x^2}$$

At $x = \frac{2A}{3}$, the initial speed is

$$v_1 = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2}$$ $$= \omega \sqrt{A^2 - \frac{4A^2}{9}}$$ $$= \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3}$$

When the speed is tripled,

$$v_2 = 3v_1 = 3 \times \omega \frac{\sqrt{5}A}{3} = \omega \sqrt{5}A$$

Let the new amplitude be $A' = \frac{nA}{3}$. Then at the same displacement,

$$v_2 = \omega \sqrt{A'^2 - x^2}$$

So,

$$\omega \sqrt{5}A = \omega \sqrt{\left(\frac{nA}{3}\right)^2 - \left(\frac{2A}{3}\right)^2}$$ $$\omega \sqrt{5}A = \omega A \sqrt{\frac{n^2 - 4}{9}}$$

Cancelling common terms,

$$\sqrt{5} = \frac{\sqrt{n^2 - 4}}{3}$$ $$3\sqrt{5} = \sqrt{n^2 - 4}$$

Squaring both sides,

$$45 = n^2 - 4$$ $$n^2 = 49$$

Since $n$ is positive,

$$n = 7$$

Therefore, the correct option is A.

Given: Initial amplitude is $A$, displacement is $\frac{2A}{3}$, and the speed becomes three times.

Find: The value of $n$ if the new amplitude is $\frac{nA}{3}$.

At $x = \frac{2A}{3}$, initial speed is

$$v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \frac{\sqrt{5}A}{3}$$

So the new speed is

$$v' = 3v = \omega \sqrt{5}A$$

Using the same SHM speed relation with new amplitude $A'$,

$$\omega \sqrt{5}A = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2}$$

Squaring,

$$5A^2 = (A')^2 - \frac{4A^2}{9}$$ $$(A')^2 = 5A^2 + \frac{4A^2}{9} = \frac{49A^2}{9}$$ $$A' = \frac{7A}{3}$$

Hence $n = 7$, so the correct option is A.