A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If the mass of the object is 0.20

Question

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If the mass of the object is 0.20 kg, what is the spring constant

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Neala 4 weeks 2021-08-17T12:53:14+00:00 1 Answers 0 views 0

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    2021-08-17T12:54:58+00:00

    Answer:

    The spring has a constant of 126.334 newtons per meter.

    Explanation:

    Given that mass-spring system experiment a simple harmonic motion, the angular frequency of the system as a function of frequency is:

    \omega = 2\pi \cdot f

    Where:

    \omega – Angular frequency, measured in radians per second.

    f – Frequency, measured in hertz.

    Given that f = 4\,hz, the angular frequency of the system is:

    \omega = 2\pi \cdot (4\,hz)

    \omega \approx 25.133\,\frac{rad}{s}

    Now, the angular frequency can be obtained in terms of spring constant and mass. That is:

    \omega = \sqrt{\frac{k}{m} }

    Where:

    k – Spring constant, measured in newtons per meter.

    m – Mass, measured in kilograms.

    The spring constant is now cleared:

    k = \omega^{2}\cdot m

    If \omega = 25.133\,\frac{rad}{s} and m = 0.20\,kg, the spring constant is:

    k = \left(25.133\,\frac{rad}{s} \right)^{2}\cdot (0.20\,kg)

    k = 126.334\,\frac{N}{m}

    The spring has a constant of 126.334 newtons per meter.

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