A mass attached to a spring oscillates with a period of 44 sec. After 33 kg are​ added, the period becomes 55 sec. Assuming that we can negl

Question

A mass attached to a spring oscillates with a period of 44 sec. After 33 kg are​ added, the period becomes 55 sec. Assuming that we can neglect any damping or external​ forces, determine how much mass was originally attached to the spring.

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Genesis 2 weeks 2021-09-13T11:31:57+00:00 2 Answers 0

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    0
    2021-09-13T11:33:13+00:00

    Step-by-step explanation:

    Below is an attachment containing the solution.

    0
    2021-09-13T11:33:46+00:00

    Answer:

    The mass originally attached to the spring was 58.665 kg

    Step-by-step explanation:

    To solve for this we need to consider the equation for a springs oscillating period:

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

    Here, m = mass of spring

    k = spring constant

    Since we know that the period increases by 11 (55 – 44) after the extra weight is added, we have the following equations:

    2*\pi *\sqrt{\frac{m}{k} }=44    -Equation 1

    2*\pi *\sqrt{\frac{m+33}{k} }=55    -Equation 2

    solving both equation simultaneously we get:

    m =  58.665 kg

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45:7+7-4:2-5:5*4+35:2 =? ( )