Question

A mass-spring system can oscillate once every 2.5s when a 750 grammass is attached to it. What must the spring constant be to allow thisvibrational period?

Answer 1

Answer:

4.737 N/m

Explanation:

The period of a mass-spring system is calculated as

$T=2\pi\sqrt{\frac{m}{k}}$

Where m is the mass and k is the spring constant. Solving for k, we get

$\begin{gathered} \frac{T}{2\pi}=\sqrt{\frac{m}{k}} \\ \\ (\frac{T}{2\pi})^2=\frac{m}{k} \\ \\ \frac{T^2}{4\pi^2}\cdot k=m \\ \\ T^2\cdot k=m(4\pi^2) \\ \\ k=\frac{m(4\pi^2)}{T^2} \end{gathered}$

Then, replacing m = 750 g = 0.75 kg and T = 2.5s, we get:

$k=\frac{0.75\text{ kg }\cdot4\cdot\pi^2}{(2.5\text{ s\rparen}^2}=4.737\text{ N/m}$

Therefore, the spring constant must be 4.737 N/m