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26 September, 18:01

An undamped 2.65 kg horizontal spring oscillator has a spring constant of 38.5 N/m. While oscillating, it is found to have a speed of 2.92 m/s as it passes through its equilibrium position. What is its amplitude of oscillation?

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  1. 26 September, 21:05
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    Answer: 0.44831m

    Explanation:

    Unwanted horizontal spring oscillator=2.65kg

    Spring constant = 38.5N/M

    Speed=2.92m/s

    Amplitude of oscillation=?

    Potential energy=m*v2/2

    =2.65*2.92/2

    =3.8695J

    Potential energy=kinetic energy

    Potential energy=1/2kx^2

    3.869=1/2*38.5*x^2

    3.869=19.25x^2

    Dividing both sides by 19.25

    3.869/19.25=x^2

    So therefore, x^2=√0.200987

    x=0.44831m
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