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    Question

    In a spring-block system executing SHM, at displacement

    x= A/2, the ratio of kinetic to potential energy is:
    A 3 Correct Answer Incorrect Answer
    B 2 Correct Answer Incorrect Answer
    C 1 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer

    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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