Question

The potential energy of a simple harmonic oscillator of mass 2 kg in its mean position is 5j. (a) if its total energy is 9j and its amplitude is 0.01m,its time period would be

Answer 1

Answer : Time period = π/100 s

Solution :
_________

answer : π/100 sec
explanation : amplitude, A = 0.01m
mass of oscillator , m = 2 kg
potential energy at mean position, P.E = 5J
total energy at mean position , T.E = 9J
at mean position,
total energy = potential energy + kinetic energy
9J = 5J + kinetic energy
kinetic energy = 4J
or, 1/2 mv² = 4J
or, 1/2 × 2 × v² = 4
or, v² = 4
hence, v = 2m/s , it is maximum speed of oscillator.
so,v=wA
2 =w × 0.01
or, 200 = w

we know,
so,

hence, time period is π/100 se