Question

The period of simple harmonic motion of mass weighing 8 pounds attached to a spring whose constant is 6.25 lb/ft is _________ seconds

Short answer

The period of simple harmonic motion of mass weighing 8 pounds attached to a spring whose constant is $6.25lb/ft$is $T=\frac{2\pi }{5}$seconds

Step-by-step solution

Definition of spring / mass system

The period of whatever object moving in a simple harmonic motion can generally be determined using the spring-mass system.

A spring/mass system's free undamped motion is represented by the equation

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0where{\omega }^2=\frac{k}{m}$

Step 2:

The period is given by $T=\frac{2\pi }{\omega }$

We were provided the following facts in this case:

The mass weight$W=8$pounds and the spring constant$k=6.25lb/ft$

From this, we get

localid="1663949576829" $$\begin{gathered} m=\frac{W}{g} \\ m=\frac{8 lb}{32 ft/s^2} \\ m=\frac{1}{4} \end{gathered}$$

We have

$$\begin{gathered} T=\frac{2\pi \sqrt{m}}{\sqrt{k}} \\ T=\frac{2\pi }{\sqrt{4·6.25}} \\ T=\frac{2\pi }{5} s \end{gathered}$$

Therefore, the period of simple harmonic motion of mass weighing 8 pounds attached to a spring whose constant is$6.25lb/ft$ is $T=\frac{2\pi }{5}$seconds