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

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 $a = 2 \times 10^{-2}\, m = 2 \,cm$
At $t =0$,
$X = 2\, cm$
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, $\omega = \frac{2\pi}{T} = \pi$
$\Rightarrow T = 2\,s$
So, required time $ = \frac{T}{4} $
$= \frac{2}{4} = 0.5\,s$