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

4.737 N/m

Step-by-step explanation:

The period of a mass-spring system is calculated as

`T=2\pi\sqrt{(m)/(k)}`

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

`\begin{gathered} (T)/(2\pi)=\sqrt{(m)/(k)} \\ \\ ((T)/(2\pi))^2=(m)/(k) \\ \\ (T^2)/(4\pi^2)\cdot k=m \\ \\ T^2\cdot k=m(4\pi^2) \\ \\ k=(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