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. T₁<T₂
• B. T₁>T₂
• C. T₁=T₂
• D. T₁=2T₂

Answer 1

The correct option is A

T₁<T₂

Using x = A sin $\omega$t

For x = $\frac{A}{2}$ , sin $\omega T_1=\frac{1}{2}\Rightarrow T_1=\frac{\pi }{6\omega }$

For x = A , sin $\omega (T_1+T_2)=1\Rightarrow T_1+T_2=\frac{\pi }{2\omega }$

$\Rightarrow T_2=\frac{\pi }{2\omega }-T_1=\frac{\pi }{2\omega }-\frac{\pi }{6\omega }=\frac{\pi }{3\omega }i.e{T}_1<T_2$

Alternate method: In S.H.M., velocity of particle also oscillates simple harmonically.

Speed is more near the mean position and less near the extreme position.

Therefore the time taken for the particle to go from 0 to $\frac{A}{2}$ will be less than

the time taken to go from $\frac{A}{2}$ to A. Hence $T_1<T_2$ .