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A mass m is attached to a spring of constant k and is undergoing SHM. The total energy of the system is proportional to:

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Answer

(2) and (4)

Step-by-step solution

8 steps

  1. Displacement in SHM

    x=Acos⁡(ωt)x = A \cos(\omega t)
    where ω=km\omega = \sqrt{\frac{k}{m}}.

  2. Velocity

    v=dxdt=−Aωsin⁡(ωt)v = \frac{dx}{dt} = -A \omega \sin(\omega t).

  3. Kinetic energy

    K=12mv2=12m(Aωsin⁡(ωt))2=12mω2A2sin⁡2(ωt).K = \frac{1}{2} m v^{2} = \frac{1}{2} m (A \omega \sin(\omega t))^{2} = \frac{1}{2} m \omega^{2} A^{2} \sin^{2}(\omega t).

  4. Substitute ω2=km\omega^{2} = \frac{k}{m}

    K=12m(km)A2sin⁡2(ωt)=12kA2sin⁡2(ωt).K = \frac{1}{2} m \left(\frac{k}{m}\right) A^{2} \sin^{2}(\omega t) = \frac{1}{2} k A^{2} \sin^{2}(\omega t).

  5. Potential energy

    U=12kx2=12k(Acos⁡(ωt))2=12kA2cos⁡2(ωt).U = \frac{1}{2} k x^{2} = \frac{1}{2} k (A \cos(\omega t))^{2} = \frac{1}{2} k A^{2} \cos^{2}(\omega t).

  6. Total energy

    E=K+U=12kA2sin⁡2(ωt)+12kA2cos⁡2(ωt)=12kA2(sin⁡2(ωt)+cos⁡2(ωt)).E = K + U = \frac{1}{2} k A^{2} \sin^{2}(\omega t) + \frac{1}{2} k A^{2} \cos^{2}(\omega t) = \frac{1}{2} k A^{2} (\sin^{2}(\omega t) + \cos^{2}(\omega t)).

  7. Simplify using trig identity sin⁡2θ+cos⁡2θ=1\sin^{2} \theta + \cos^{2} \theta = 1

    E=12kA2.E = \frac{1}{2} k A^{2}.

  8. Proportionality

    E∝kE \propto k and E∝A2E \propto A^{2}.
    EE does not depend on mm, and is not proportional to AA.

    Final answer: (2)and(4)(2) and (4)

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