Answer to Question #165836 in Physics for Jonathan

Question

In the laboratory, you attach a 300 g object to a spring of negligible mass and start oscillating it. The elapse time from when the object first moves through the equilibrium position to the second time it moves through that is 2 seconds. Find the force constant of the spring.

Answer 1

The time that the object takes to start from the equilibrium position to the maximum amplitude and back to the equilibrium position is the half period of the oscillation:

$$t=\dfrac{T}{2},$$ $$T=2t=2\cdot2\ s=4\ s.$$

Let's find the angular frequency of the oscillation:

$$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{4\ s}=1.57\ \dfrac{rad}{s}.$$

Finally, we can find the force constant of the spring:

$$\omega=\sqrt{\dfrac{k}{m}},$$ $$k=\omega^2m=(1.57\ \dfrac{rad}{s})^2\cdot0.3\ kg=0.74\ \dfrac{N}{m}.$$