A) 3/4
B) zero
C) 1/2
D) 1/4
Correct Answer: A
Solution :
The total energy in SHM\[=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] The kinetic energy\[=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] Given: \[x=\frac{A}{2}\] \[\therefore \] \[KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-\frac{{{A}^{2}}}{4} \right)\] \[KE=\frac{3}{4}\]total energy
You need to login to perform this action.
You will be redirected in
3 sec
