JEE Physics 2023 Question with Solution

Given: The particle is in SHM between $-A$ and $A$. The time from $x=0$ to $x=A$ is stated as $2 \, \text{s}$. The solution working uses the intermediate motion from $0$ to $A/2$ as $t_1$ and from $A/2$ to $A$ as $t_2$.

Find: The required time according to the extracted solution working.

Using the SHM form

$$x = A \sin(\omega t)$$

when the particle reaches $A/2$,

$$\frac{A}{2} = A \sin(\omega t_1)$$

so,

$$\sin(\omega t_1) = \frac{1}{2}$$

Hence,

$$\omega t_1 = \frac{\pi}{6}$$

For motion from $A/2$ to $A$, let the additional time be $t_2$. At $x=A$,

$$A = A \sin\big(\omega (t_1+t_2)\big)$$

therefore,

$$\sin\big(\omega (t_1+t_2)\big)=1$$

which gives

$$\omega (t_1+t_2)=\frac{\pi}{2}$$

So,

$$\omega t_2 = \frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$$

Thus,

$$\frac{t_2}{t_1}=\frac{\pi/3}{\pi/6}=2$$

and hence,

$$t_2 = 2t_1$$

The extracted solution then substitutes $t_1 = 2 \, \text{s}$ and gets

$$t_2 = 2 \times 2 = 4 \, \text{s}$$

Therefore, the extracted solution declares the correct option is C. There is a discrepancy because option C is $1.5 \, \text{s}$ while $4 \, \text{s}$ corresponds to option D.

Given: the solution contains two inconsistent claims: one header says option C, while the numerical working gives $4 \, \text{s}$.

Find: Which answer should be taken from the solution.

The numerical derivation ends with

$$t_2 = 4 \, \text{s}$$

The options are:

• A = $3 \, \text{s}$
• B = $2 \, \text{s}$
• C = $1.5 \, \text{s}$
• D = $4 \, \text{s}$

Since the working concludes $4 \, \text{s}$, that value matches option D. However, the solution explicitly labels the correct option as C. Following the instruction to prioritize the solution, the answer is taken as the option label concluded on the page, while noting the mismatch with the numerical value.