Question

This displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale)

Which of the following statements is/are true for this motion?

• A. (B), (C) and (D)
• B. (A), (B) and (D)
• C. (A) and (D)
• D. (A), (B) and (C)

Answer 1

The correct option is D

(A), (B) and (C)

Step 1: Explanation for (A)

at $t=\frac{3T}{4}$

The particle is at the mean position, means no displacement

Therefore, $a$ = 0 and F = 0. ($F$ is the force on the oscillator and $a$ is the acceleration of it.)

Step 2: Explanation for (B)

at $t=T$,

The particle is at an extreme position.

Therefore, F is maximum and $a$ is also maximum.

Step 3: Explanation for (C)

at $t=\frac{T}{4}$ ; mean position

Therefore, velocity is maximum.

Step 4: Explanation for (D)

At, $t=\frac{T}{2}$, K.E. = P.E.

Therefore,

$\frac{1}{2}k\left(A^2-x^2\right)=\frac{1}{2}k{x}^2$

$\Rightarrow \left(A^2-x^2\right)=x^2$

$$\begin{gathered} \Rightarrow A^2=2x^2 \\ \Rightarrow A=\sqrt{2x^2} \\ \Rightarrow x=\frac{A}{\sqrt{2}} \end{gathered}$$

$\Rightarrow x=A\cos \omega t,Where,\left(Cos\omega t=\frac{1}{\sqrt{2}}\right)$

Therefore,

$$\begin{gathered} \omega t=\frac{\pi }{4} \\ So,\frac{2\pi }{T}.t=\frac{\pi }{4} \\ \Rightarrow t=\frac{T}{8} \end{gathered}$$

Hence, the correct option is (D).