Question
A linear harmonic oscillator of force constant $$2 \times {10^6}N/m$$ and amplitude $$0.01\,m$$ has a total mechanical energy of $$160\,J.$$ Its
A.
maximum potential energy is $$160\,J$$
B.
maximum potential energy is $$100\,J$$
C.
maximum potential energy is zero
D.
minimum potential energy is $$100\,J$$
Answer :
maximum potential energy is $$160\,J$$
Solution :
The potential energy of a particle executing $$SHM$$ is given by,
$$U = \frac{1}{2}m{\omega ^2}{x^2}$$
$$U$$ is maximum, when $$x = a =$$ amplitude of vibration i.e. the particle is passing from the extreme position and is minimum when $$x = 0,$$ i.e. the particle is passing from the mean position
$$\therefore {U_{\max }} = \frac{1}{2}m{\omega ^2}{a^2}\,......\left( {\text{i}} \right)$$
Also, total energy of the particle at instant $$t$$ is given by
$$E = \frac{1}{2}m{\omega ^2}{a^2}\,......\left( {{\text{ii}}} \right)$$
So, when $$E = 160\,J,$$ then maximum potential energy of particle will also be $$160\,J.$$
Alternative
$${\left( {KE} \right)_{\max }} = {\left( {PE} \right)_{\max }} = {\text{Total Mechanical Energy}}$$
So, Total Mechanical Energy $$= 160\,J$$