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. $0.5s$
• B. $0.75s$
• C. $0.125s$
• D. $0.25s$

Answer 1

Answer: (A)

$x=\left(2\times 10^{-2}\right)\cos \pi t$
Here, $a=2\times 10^{-2}m=2cm$
At $t=0,x=2cm$ , $ie$ , 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