The displacement of a particle, executing simple harmonic motion with time period T, is expressed as x(t) = A t, where A is the amplitude. The maximum value of potential energy of this oscillator is found at t = T/2 . The value of is .
Answer
2
Solution
U = 12kx2 is maximum when |x| = A, i.e. t = 1.
t = 2
2 Tt = 2
t = T4 (first instant)
t = T2
2 = 4, = 2
Answer: 2