Question

A particle starts from point $x=-\frac{\sqrt{3}}{2}A$ and move towards negative extreme as shown in the figure below. If the time period of oscillation is $T$, then

• A. $\text{The equation of the SHM is}x=A\sin (\frac{2\pi }{T}t+\frac{4\pi }{3})$.
• B. $\text{The time taken by the particle to go directly from its initial position to negative extreme is}\frac{T}{12}$.
• C. $\text{The time taken by the particle to reach mean position is}\frac{T}{3}$.
• D. $\text{The equation of the SHM is}x=A\sin (\frac{2\pi }{T}t+\frac{\pi }{3})$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\text{The equation of the SHM is}x=A\sin (\frac{2\pi }{T}t+\frac{4\pi }{3})$.
B $\text{The time taken by the particle to go directly from its initial position to negative extreme is}\frac{T}{12}$.
C $\text{The time taken by the particle to reach mean position is}\frac{T}{3}$.

Figure shows the solution of the problem with the help of phasor diagram.

Horizontal component of velocity at Q gives the required direction of velocity at $t=0$.


In $\Delta OSQ,\cos \theta =\frac{\frac{\sqrt{3}A}{2}}{A}=\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{6}$

From the figure, we can deduce the initial phase of the particle as
$\varphi =\frac{3\pi }{2}-\frac{\pi }{6}=\frac{8\pi }{6}=\frac{4\pi }{3}$
So, equation of SHM is $x=A\sin (\omega t+\frac{4\pi }{3})$
Now, the particle to reach the left extreme point, it will travel angle $\theta$ along the circle. So time taken.
$t=\frac{\theta }{\omega }=\frac{\pi }{6\omega }\Rightarrow t=\frac{T}{12}\text{s}$
To reach the mean position, the particle will travel an angle
$\alpha =\frac{\pi }{2}+\frac{\pi }{6}=\frac{2\pi }{3}$
So, time taken $=\frac{\alpha }{\omega }=\frac{T}{3}\text{s}$
Thus, options (a) , (b) and (c) are the correct answers.