Question

A particle of mass m moves with the potential energy U shown above.The period of the motion when the particle has total energy E is

• A. $2\pi \sqrt{m/k}+4\sqrt{2E/m{g}^2}$
• B. $2\pi \sqrt{m/k}$
• C. $\pi \sqrt{m/k}+2\sqrt{2E/m{g}^2}$
• D. $2\sqrt{2E/m{g}^2}$

Answer 1

Answer: (C)

$\pi \sqrt{m/k}+2\sqrt{2E/m{g}^2}$

As seen from graph, we can infer that for $x<0$ particle is in SHM as in spring. and for $x>0$ particle is in influence of gravity i.e. thrown upwards with some initial velocity.
So we can divide the time period in two parts. first $T_1$ and second $T_2$
$T_1$ is half of the time period of a full SHM i.e. $T_1=\pi \sqrt{\frac{m}{k}}$
$E=\frac{1}{2}m{v_{max}}^2\Rightarrow v_{max}=\sqrt{\frac{2E}{m}}$
and $T_2$ is the time during which the particle remains in air when it is thrown upwards with velocity $v_{max}=\sqrt{\frac{2E}{m}}$
$0=\sqrt{\frac{2E}{m}}t-\frac{1}{2}g{t}^2$ Since $s=ut-\frac{1}{2}g{t}^2$
$\Rightarrow T_2=2\sqrt{2E/m{g}^2}$
So total time time period $T=\pi \sqrt{\frac{m}{k}}+2\sqrt{\frac{2E}{m{g}^2}}$