Question

A particle performing simple harmonic motion undergoes initial displacement of $A/2$ (where $A$ is the amplitude of simple harmonic motion) in $1s$ . At $t=0$ , the particle may at the extreme position (or) mean position . The time period of the simple harmonic motion can be

• A. $6s$
• B. $2.4s$
• C. $12s$
• D. $1.2s$

Answer 1

Answer: (A), (C)

The correct options are
A $6s$
C $12s$

AT $t=0$ when particle is at extreme position, the situation is as shown in fig.
From the figure, $\cos \theta =\frac{A/2}{A}=\frac{1}{2}$
$\theta =\frac{\pi }{3}\Rightarrow \frac{\pi }{3}=\frac{2\pi }{T}\times 1\Rightarrow T=6s$
At $t=0$ when particle is at mean position, the situation is as shown in figure.
From the figure, $\sin \theta =\frac{A/2}{A}$
$\theta =\frac{\pi }{6}$, but $\theta =\omega t\Rightarrow \frac{\pi }{6}=\frac{2\pi }{T}\times 1\Rightarrow T=12s$
If initially the particle is located somewhere else, then time period comes out to be different. A reverse question can also be formed on the same concept.