A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does the energy of

Question

A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does the energy of the mass/spring system decrease over that time?

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Calantha 5 months 2021-08-15T01:14:53+00:00 1 Answers 1 views 0

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    2021-08-15T01:16:18+00:00

    Answer:

    The correct answer is “0.246“.

    Explanation:

    Given that the amplitude is decreased by a factor of 9, then

    A \rightarrow (A-\frac{A}{9} )

    A \rightarrow \frac{8A}{9}

    As we know,

    Energy will be:

    ⇒  E_{1}=\frac{1}{2}KA^2

    and,

    ⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

              =\frac{64KA^2}{162}

    ⇒  \Delta E=E_1-E_2

    On putting the estimated values, we get

               =\frac{1}{2}KA^2-\frac{64KA^2}{162}

    ⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

              =\frac{40}{162}

              =0.246

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