Question

A particle executes simple harmonic motion between $x=-A$ and $x=+A$. The time taken for it to go from $O$ 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 $T_1<T_2$

Since time period is only dependent on the length of the pendulum and acceleration due to gravity. So it will remain same. $\Rightarrow$B incorrect.
Since potential energy is only dependent on the displacement from mean position, it will remain as same as previous. $\Rightarrow$C incorrect.
Kinetic energy at mean position = total energy $=\frac{1}{2}k{A}^2$ (this much kinetic energy is added to pendulum)
So the total energy becomes $\frac{1}{2}k{A}^2+\frac{1}{2}k{A}^2=k{A}^2=\frac{1}{2}k{\left(\sqrt{2}A\right)}^2=\frac{1}{2}k{A^{\prime }}^2$
$\Rightarrow A^{\prime }=\sqrt{2}A$
Its new amplitude is $\sqrt{2}$ times of previous amplitude. $\Rightarrow$A correct