I attach a 4.1 kg block to a spring that obeys Hooke’s law and supply 3.8 J of energy to stretch the spring. I release the block and it osci

Question

I attach a 4.1 kg block to a spring that obeys Hooke’s law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

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Thu Cúc 5 months 2021-08-17T11:56:24+00:00 1 Answers 0 views 0

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    2021-08-17T11:58:08+00:00

    Answer:

    The amplitude of the oscillation is 2.82 cm

    Explanation:

    Given;

    mass of attached block, m = 4.1 kg

    energy of the stretched spring, E = 3.8 J

    period of oscillation, T = 0.13 s

    First, determine the spring constant, k;

    T = 2\pi \sqrt{\frac{m}{k} }

    where;

    T is the period oscillation

    m is mass of the spring

    k is the spring constant

    T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\

    Now, determine the amplitude of oscillation, A;

    E = \frac{1}{2} kA^2

    where;

    E is the energy of the spring

    k is the spring constant

    A is the amplitude of the oscillation

    E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A =  \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm

    Therefore, the amplitude of the oscillation is 2.82 cm

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