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. $0.25s$
• B. $0.50s$
• C. $0.75s$
• D. $0.125s$

Answer 1

Answer: (B)

To determine the position velocity, etc at first we write the general representation of wave and then compare the given wave equation with general wave equation
Given $X=(2\times 10^{-2})\cos \pi t$
This gives $a=2\times 10^{-2}m=2cm$
At $t=0$,
$X=2cm$
i.e. the object is at positive extreme, so to acquire maximum speed (i.e. the 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, required time $=\frac{T}{4}$
$=\frac{2}{4}=0.5s$