A) kinetic energy is minimum, potential energy is maximum
B) both kinetic and potential energies are maximum
C) kinetic energy is maximum, potential energy is minimum
D) both kinetic and potential energies are minimum
Correct Answer: C
Solution :
Kinetic energy of a particle of mass m in SHM at any point \[=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\] and potential energy \[=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] where, a is amplitude of particle and \[x\] is the distance from mean position. So, at mean position, \[x=0\] \[\therefore \] \[KE=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] (maximum) and \[PE=0\] (minimum)
You need to login to perform this action.
You will be redirected in
3 sec
