Answer to Question #146464 in Physics for Kylie

Question

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.

Answer 1

By Hook's law, the spring constant is:

$$k = \dfrac{F}{\Delta y} = \dfrac{mg}{\Delta y}$$

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

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

The frequency of the spring oscillatior is given by:

$$\nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} = \dfrac{1}{2\pi}\sqrt{\dfrac{49}{1}} = 3.5\pi\approx 11Hz$$

Answer. 11 Hz.