Question

1. (4pts) a mass, m, is oscillating on a spring of spring constant k with an angular frequency of ω. initially the period of the motion is t = 10 s. then, the spring constant is increased by a factor of 50, k → 50k, and the mass is doubled, m → 2m, what will be the new period of the motion?

Explanation:

Step1: Recall the formula for the period of a mass - spring system

The period of a mass - spring system is given by \(T = 2\pi\sqrt{\frac{m}{k}}\).

Step2: Find the initial relationship

Initially, \(T_1=10\ s = 2\pi\sqrt{\frac{m}{k}}\).

Step3: Find the new period formula

After the changes, \(m
ightarrow2m\) and \(k
ightarrow50k\). The new period \(T_2 = 2\pi\sqrt{\frac{2m}{50k}}\).

Step4: Simplify the new period formula

\(T_2=2\pi\sqrt{\frac{2m}{50k}}=2\pi\sqrt{\frac{m}{25k}}=\frac{1}{5}\times2\pi\sqrt{\frac{m}{k}}\).

Step5: Substitute the initial period value

Since \(T_1 = 2\pi\sqrt{\frac{m}{k}} = 10\ s\), then \(T_2=\frac{1}{5}T_1\).

Answer:

\(2\ s\)