A mass on a string of unknown length oscillates as a pendulum with a period of 4.00 s. What is the period if a. The mass is doubled? b. The

Question

A mass on a string of unknown length oscillates as a pendulum with a period of 4.00 s. What is the period if a. The mass is doubled? b. The string length is doubled? c. The string length is halved? d. The amplitude is halved? Parts a to d are independent questions, each referring to the initial situation.?

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Thiên Thanh 3 years 2021-08-26T11:41:51+00:00 2 Answers 25 views 0

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  1. Orla
    0
    2021-08-26T11:43:16+00:00
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    Explanation:

    Below are attachments containing the solution.

  2. thuthao
    0
    2021-08-26T11:43:20+00:00
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    Answer:

    a) T= 4.00s

    b)T = 5.66s

    c)T  = 2.83s

    d)T = 4.00s

    Explanation:

    a)  The pendulum period is express as

    T_0 = 2\pi \sqrt{\frac{L_0}{g} } = 4.00s

    The period does not depend on the mass and depend only on the length

    so we have,

    T = T_0 = 4.00s

    b) for new length

    L = 2L_0

    so, we have

    T_0 = 2\pi \sqrt{\frac{L_0}{g} } = \sqrt{2T_0} = 5.66s

    c) For a new length

    T_0 = 2\pi \sqrt{\frac{L_0}{g} } = \frac{1}{\sqrt{2} } T_0= 2.83s

    d) the period does not depend on the amplitude as long as there is simple harmonic motion

    so, we have

    T = 4.00s

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )