Question

A particle performs SHM with a frequency of $1/2$ Hz. If the time is measured from its instantaneous rest position, then the new frequency (in Hz) when its displacement becomes equal to half of its amplitude is:

• A. $1/12$
• B. $1/6$
• C. $1/24$
  
• D. $1/2$

Answer 1

Answer: (D)

Frequency $f=\frac{1}{2}Hz$

Angular frequency $\omega =2\pi f=2\pi \times \frac{1}{2}=\pi$

If A is amplitude its displacement

$x=A\sin \omega t$

or $x=A\sin \pi t$

Now for $t=t_1$ $x=\frac{A}{2}$

$\therefore \frac{A}{2}=A\sin \pi t_1$

$\Rightarrow \sin \pi t_1=\frac{1}{2}$

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

$\Rightarrow t_1=\frac{1}{6}$ sec

This is time from mean position

Its instantaneous rest position is at $x=A$,

So, $A=A\sin \pi t_2$

$\Rightarrow t_2=\frac{1}{2}$ sec