Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.31 m and finds that it makes 247 complete osc

Question

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.31 m and finds that it makes 247 complete oscillations in 793 s. The amplitude of the oscillations is very small compared to the pendulum’s length. What is the gravitational acceleration on the surface of this planet?

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Thái Dương 3 years 2021-07-22T04:19:25+00:00 1 Answers 39 views 0

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  1. kiet
    0
    2021-07-22T04:20:44+00:00
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    Answer:

    The gravitational acceleration is 5.03 m/s² on the surface of the planet.

    Explanation:

    Frequency:

    The number of complete oscillation per unit second is called frequency.

    frequency=\frac{\textrm {Number of oscillation}}{Time}

    Angular frequency:

    Angular frequency is the product of the oscillation frequency and the angle thought which the object moves.

    Angular velocity (\omega)=2\pi f

    Given that,

    Number of oscillation= 247, time= 793 s, length of the pendulum=1.31 m

    f=\frac{\textrm {Number of oscillation}}{Time}

       =\frac{247}{793} Hz.

    Angular velocity

    \omega=2\pi f

       =2 \times \pi \times \frac{247}{793}

      =1.96 rad/s

    We know that,

    g=\omega^2l

    g= gravitational force of the planet

    l = length of the pendulum

    \omega = angular velocity

    g=(1.96)^2\times 1.31 m/s²

      = 5.03 m/s²

    The gravitational acceleration on the surface of the planet is 5.03 m/s².

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