Question

For the spring pendulum shown in fig. the value of spring constant is `3xx10^4(N)/(m)` and amplitude of oscillation is `0.1m`. The total mechanical energy of oscillating system is 200 J. Mark out the correct option (s).
A. Minimum PE of the oscillating system is 50 J
B. Maximum PE of the oscillating system is 200J
C. maximum KE of the oscillating system is 200 J
D. minimum KE of the oscillating system is 150 J.

Answer 1

Correct Answer - A::B
Total mechanical energy of the oscillating system is `E=K_(max)+U_(min)+K_(min)`,`K_(min)=0` at extreme position.
So, `U_(max)=E=200J`
`K_(max)=(mv_(max)^2)/(2)=(mxxA^2omega^2)/(2)=(KA^2)/(2)=150J`
So, `U_(min)=50J`