Question #165836
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.
1
Expert's answer
2021-02-23T10:04:44-0500

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=T2,t=\dfrac{T}{2},T=2t=22 s=4 s.T=2t=2\cdot2\ s=4\ s.

Let's find the angular frequency of the oscillation:


ω=2πT=2π4 s=1.57 rads.\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:


ω=km,\omega=\sqrt{\dfrac{k}{m}},k=ω2m=(1.57 rads)20.3 kg=0.74 Nm.k=\omega^2m=(1.57\ \dfrac{rad}{s})^2\cdot0.3\ kg=0.74\ \dfrac{N}{m}.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS