Question

The amplitude of a particle in SHM is $\text{5}cms$ and its time period is $\pi .$ At a displacement of $3cm$ from its mean position the velocity in cm/sec will be:

• A. 8
• B. 12
• C. 2
• D. 16

Answer 1

Answer: (A)

8

Let equation of SHM is $x=Asin\left(\omega t\right)$, Where $A=5$ $cms$,

$\therefore v=A\omega cos\left(\omega t\right)$

Given, Time Period$=\pi =\frac{2\pi }{\omega }$ $\Rightarrow \omega =2$

$x=5sin\left(2t\right)$,

At a displacement $3$ $cm$, $\Rightarrow 3=5sin\left(2t\right)$ $\Rightarrow cos\left(2t\right)=\frac{4}{5}$,

$v=5\times 2cos\left(2t\right)=5\times 2\times \frac{4}{5}=8cm{s}^{-1}$