Question

A particle executes simple harmonic motion between x = -A and x = +A. The time taken for it to go from 0 to A/2 is $T_1$ and to go from A/2 to A is $T_2.$ Then

• A. $T_1<T_2$
• B. $T_1>T_2$
• C. $T_1=T_2$
• D. $T_1=2T_2$

Answer 1

The correct option is A $T_1<T_2$

Method 1: Qualitative. The velocity of a body executing S.H. M is maximum at its centre and decreases as the body proceeds to the extremes. Therefore if the time taken for the body to go from O to A/2 is $T_1$ and then to go A is $T_2.$ then obviously $T_1$ < $T_2.$ .
Method 2: Quantitative. Any S.H.M. is given by the equation y =a $sin\omega t$ where y is the displacement of the body at any instant t. a is the amplitude and $\omega$is the angular frequency.
When x =0, $\omega t_1=0\therefore t_1=0$

When x = a/2, $\omega t_2=\pi /6t_2=\frac{\pi }{6\omega }$
When x = a, $\omega t_3=\pi /2t_3=\frac{\pi }{2\omega }$

Time taken for O to a/2 will be $t_2-t_1=\frac{\pi }{6\omega }=T_1$

Time taken for a/2 to a will be

$t_3-t_2=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }=\frac{2\pi }{6\omega }=\frac{\pi }{3\omega }=T_2$