Question

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

Answer: (A)

$T_1<T_2$

Any SHM is given by the equation $x=sin\omega t,$ where x is the displacement of the body 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 $x=A/2,\omega t_2=\pi /6,t_2=\pi /6\omega$

When $x=A,\omega t_3=\pi /2,t_3=\pi /2\omega$

Time taken from $0$ to $A/2$ will be

$t_2-t_1=\frac{\pi }{6\omega }=T_1$

Time taken from $A/2$ to $A$ will be

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

Hence, $T_2>T_1$