Question
In a spring-block system executing SHM, at displacement
x= A/2, the ratio of kinetic to potential energy is:Solution
In a spring-block system undergoing simple harmonic motion (SHM) , the total mechanical energy E is constant and given by: E = KE + PE = (1/2) kA2 At any displacement x, the potential energy is: PE = kx2/2 And the kinetic energy is the remaining part: KE = (k/2)(A2 - x2) At x = A/2, PE = k(A/2)2/2 = kA2/8 KE = (k/2){A2 - (A/2)2} KE = (k/2)(A2 - A2/4) KE = 3kA2/8 We are asked to find the ratio KE/PE at x = A/2 
The ratio of kinetic to potential energy at displacement x=A/2 is 3:1.
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