An un-stretched vertical spring has a length of 0.50 m. When a 1.0 kg mass is attached, it stretches and comes to rest at a length of 0.70 m (Δy = 0.20 m). This spring is then stretched farther and released. This puts the system into simple harmonic motion. What is the frequency (Hz) for the oscillations? Hint: solve for spring constant k first and then solve for frequency.
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Expert's answer
2020-11-25T10:50:36-0500
By Hook's law, the spring constant is:
k=ΔyF=Δymg
where F=mg is the force applied to the spring, m=1kg is the mass of the ball, Δy=0.2m is the extension of the spring, and g=9.8m/s2. Thus, obtain:
k=0.21⋅9.8=49N/m
The frequency of the spring oscillatior is given by:
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