Question

The displacement of an object attached to a spring and executing simple harmonic motion is given by $x=2\times 10^{- 2}cos \pi t$ metre. The time at which the maximum speed first occurs is

• A. $0.5 \, s$
• B. $0.75 \, s$
• C. $0.125 \, s$
• D. $0.25 \, s$

Answer 1

Answer: (A)

$x=\left(2 \times 10^{-2}\right) \cos \pi t$
Here, $a=2\times 10^{- 2}m=2cm$
At $t=0, \, x=2 \, cm$ , $ie$ , the object is at the positive extreme, so to acquire maximum speed (i.e., to reach mean position) it takes $\frac{1}{4}$ th of the time period.
$\therefore $ Required time = $\frac{T}{4}$
where $\omega =\frac{2 \pi }{T}=\pi $
$\Rightarrow \, \, \, T=2s$
So, the required time = $\frac{T}{4}=\frac{2}{4}=0.5$ s