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 _ _ _ _ .

Step-by-step solution

For SHM with x(t) = A ( t) , the potential energy is U = 1 2 m ^2 x^2 = 1 2 m ^2 A^2 ^2( t) . Maximum potential energy occurs when ^2( t) = 1 , i.e., when t = 2 , 3 2 , ... The first maximum occurs at t = 2 , giving t = 2 = 2 T 2 = T 4 . Since this is given as t = T 2 . We have T 4 = T 2 , so = 2 .