Question

A particle executing $S.H.M$ of amplitude $4cm$ and $T=4sec$. The time taken by it move from positive extreme position to half the amplitude is:-

• A. $1sec$
• B. $\frac{1}{3}sec$
• C. $\frac{2}{3}sec$
• D. $\sqrt{\frac{2}{3}}sec$

Answer 1

The correct option is B $\frac{1}{3}sec$

If $y$ is the displacement, $t$ the time,$a$ the amplitude and $\omega$ the angular frequency

then $y=a\cos \omega t$

Given $y=\frac{a}{2}\Rightarrow \frac{a}{2}=a\cos \omega t$

$\Rightarrow \cos \omega t=\frac{1}{2}$

$\Rightarrow \frac{1}{2}=\cos \frac{2\pi }{T}\times 2$ at $T=4$secs

$\Rightarrow t\cos \frac{2\pi }{4}=\frac{1}{2}$

Let $W_A$ be the work done by the spring $A$ and $W_B$ be the work done by the spring $B$

$\Rightarrow \frac{W_A}{W_B}=\frac{K_A}{K_B}=\frac{1}{3}$

$\therefore W_A:W_B=1:3$