Question

A free undamped spring/mass system oscillates with a periodof 3 seconds. When 8 pounds are removed from the spring, thesystem has a period of 2 seconds. What was the weight of theoriginal mass on the spring?

Short answer

The weight of the original mass on the spring is $14.4\text{lbs}$.

Step-by-step solution

Step 1:Define time period.

The time period of a free undamped spring/mass system is represented by$T=\frac{2\pi }{\omega }$. Here, ${\omega }^2=\frac{k}{m}=\frac{kg}{W},k$is the spring constant, W is the weight of the mass attached to the spring and is the acceleration due to gravity. Then, the time period can be represented as$T=2\pi \sqrt{\frac{W}{Kg}}$… (1)

Determine the weight of the original mass on the spring.

Given that the system oscillates with a period of 3 seconds. Then, the equation becomes$2\pi \sqrt{\frac{W}{Kg}}=3$… (2)

Whenpounds are removed from the spring, the system has a period ofseconds, i.e.$2\pi \sqrt{\frac{W-8}{Kg}}=2$,… (3)

Divide the equation (3) by (2) to get:

$\begin{array}{c}\sqrt{\frac{W}{W-8}}=\frac{3}{2}\\ \frac{W}{W-8}=\frac{9}{4}\\ 4W=9W-72\\ W=14.4\text{lbs}\end{array}$

Hence, the weight of the original mass on the spring is$14.4\text{lbs}$.