Question

If maximum amplitude of an SHM is 2 and Time period is $\pi /8$ sec, displacement time equation of the particle is:

• A. x $=$ 2 sin 2t
• B. x $=$ 2 sin4t
• C. x $=$ 2 cos2t
• D. none of the above

Answer 1

Answer: (D)

none of the above

Displacement $x$ of any particle in SHM at time $t$ is given by $x=A\sin \omega t$

where $A$ is the amplitude, $\omega$ is the angular frequency$=\frac{2\pi }{T}$ and $T$ is the time period of oscillation.

Given : $T=\frac{\pi }{8}$ and $A=2$
Thus $\omega =\frac{2\pi }{T}=\frac{2\pi }{\pi /8}=16$
So Displacement of the particle $x=2\sin 16t$