Question

A mass weighing 4 pounds is attached to a spring whose spring constant is $\mathbf{16}\mathbf{lb}\mathbf{/}\mathbf{ft}$. What is the period of simple harmonic motion?

Short answer

So, the required solution is$T=\frac{\sqrt{2}\pi }{8}$ .

Step-by-step solution

Definition of Simple Harmonic Motion

Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position.

Solve for Mass

Suppose that, a mass weighing 4 pounds is attached to a spring whose spring constant is 16$\text{lb}/\text{ft}$.

The objective is to find the period of simple harmonic motion.

From the data, spring constant$k=16\text{lb}/\text{ft}$

Mass$m=\frac{\text{weight}}{g}$

$=\frac{4}{32}[\text{Useweight}=4,g=32]$

$=\frac{1}{8}\text{slugs}$

Period of Simple Harmonic Motion

Period of the simple harmonic motion is,

$T=\frac{2\pi }{\omega }$

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

$=\frac{2\pi }{\sqrt{16/(1/8)}}[\text{Use}k=16,m=\frac{1}{8}]$

$=\frac{2\pi }{\sqrt{8\times 16}}$

$=\frac{2\pi }{\sqrt{2\times 64}}$

$=\frac{2\pi }{8\sqrt{2}}$

$=\frac{\sqrt{2}\pi }{8}$

Therefore, period of the simple harmonic motion is, $T=\frac{\sqrt{2}\pi }{8}$.