Question

The displacement of an object attached to a spring and executing simple harmonic motion is given $byx = 2 \times 10^{-2} cos\,\pi t$ 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)

$0.50\,s$

Answer 2

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 $= \frac{T}{4}$
$= \frac{2}{4} = 0.5\,s$