Question

The time period of an object performing SHM is $16\text{s}$. It starts its motion from the equilibrium position. After $2\text{s}$ its velocity is $\pi \text{m/s}$. What is its displacement amplitude?

• A. $\sqrt{2}\text{m}$
• B. $2\sqrt{2}\text{m}$
• C. $4\sqrt{2}\text{m}$
• D. $8\sqrt{2}\text{m}$

Answer 1

The correct option is D $8\sqrt{2}\text{m}$

Displacement of the object, $x=\text{A sin}\omega t$
Velocity, $v=A\omega \cos \omega t$
Since, $\omega =\frac{2\pi }{T}=\frac{2\pi }{16}$
Therefore, $v=A\left(\frac{2\pi }{16}\right)\text{cos}\frac{2\pi }{16}\times 2$
$\pi =A\times \frac{2\pi }{16}\times \frac{1}{\sqrt{2}}$
Hence, $A=\pi \times \frac{16}{2\pi }\times \sqrt{2}=8\sqrt{2}\text{m}$