Question

A particle executes simple harmonic motion (amplitude $=A$) between $x=-A$ and $x=+A$. The time taken for it to go from $0$ to $\frac{A}{2}$ is $T_1$ and to go from $\frac{A}{2}$ 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 $T_1<T_2$

Using $x=A\text{sin}\omega t$
For $x=\frac{A}{2},\text{sin}\omega T_1=\frac{1}{2}\Rightarrow T_1=\frac{\pi }{6\omega }$
For $x=A,\text{sin}\omega (T_1+T_2)=1\Rightarrow T_1+T_2=\frac{\pi }{2\omega }$
$\Rightarrow T_2=\frac{\pi }{2\omega }-T_1=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }=\frac{\pi }{3\omega }\text{i.e.}{T}_1<T_2$