Answer to Question #146464 in Physics for Kylie

Question #146464
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.
1
Expert's answer
2020-11-25T10:50:36-0500

By Hook's law, the spring constant is:


k=FΔy=mgΔyk = \dfrac{F}{\Delta y} = \dfrac{mg}{\Delta y}

where F=mgF = mg is the force applied to the spring, m=1kgm = 1kg is the mass of the ball, Δy=0.2m\Delta y = 0.2m is the extension of the spring, and g=9.8m/s2g = 9.8m/s^2. Thus, obtain:


k=19.80.2=49N/mk = \dfrac{1\cdot 9.8}{0.2} =49N/m

The frequency of the spring oscillatior is given by:


ν=12πkm=12π491=3.5π11Hz\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi}\sqrt{\dfrac{49}{1}} = 3.5\pi\approx 11Hz

Answer. 11 Hz.


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