A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the perio

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A simple pendulum is made my attaching a rod of negligible mass to a 2.0 kg pendulum bob at the end. It is observed that on Earth, the period of small-angle oscillations is 1.0 second. It is also observed that on Planet X this same pendulum has a period of 1.8 seconds. How much does the pendulum bob weigh on Planet X

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Thu Thảo 4 years 2021-08-02T20:08:43+00:00 1 Answers 31 views 0

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  1. MichaelMet
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    2021-08-02T20:10:16+00:00
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    Answer:

    W = 9.081\,N

    Explanation:

    The angular frequency of a simple pendulum is:

    \omega = \sqrt{\frac{g}{l} }

    Where:

    g – Gravitational constant, in \frac{m}{s^{2}}.

    l – The rod length, in m.

    The period of oscillation of the simple pendulum is:

    T = 2\pi\cdot \sqrt{\frac{l}{g} }

    Given that the same pendulum is tested of both planets, the following relation is determined:

    T_{1}^{2}\cdot g_{1} = T_{2}^{2}\cdot g_{2}

    The gravity constant on Planet X is:

    g_{2} = g_{1}\cdot \left(\frac{T_{1}}{T_{2}}\right)^{2}

    g_{2} = (9.807\,\frac{m}{s^{2}})\cdot \left(\frac{1\,s}{1.8\,s} \right)^{2}

    g_{2} = 3.027\,\frac{m}{s^{2}}

    The weight of the pendulum bob on Planet X is:

    W = (3\,kg) \cdot (3.027\,\frac{m}{s^{2}} )

    W = 9.081\,N

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