Question

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

• A. 1 sec
• B. 1/3 sec
• C. 2/3 sec
• D. √3/2 sec

Answer 1

The correct option is C 2/3 sec

Here positive extreme means that it is starting from highest point $(y=+a).$
Therefore, for simplifying the sum let us consider the equation
$y=acosw{t}_{(sinceatwt=0\Rightarrow y=a)}$
$\Rightarrow \frac{a}{2}=acoswt\Rightarrow coswt=\frac{1}{2}\Rightarrow wt=\frac{\pi }{3}$

$\Rightarrow \frac{2\pi t}{T}=\frac{\pi }{3}\Rightarrow t=\frac{\frac{\pi }{3}\times T}{2\pi }=\frac{4}{3\times 2}=\frac{2}{3}sec$