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 $x=Asin\omega t$
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 $T_1<T_2$