# 10.2 Simple harmonic motion: a special periodic motion  (Page 2/7)

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In fact, the mass $m$ and the force constant $k$ are the only factors that affect the period and frequency of simple harmonic motion.

## Period of simple harmonic oscillator

The period of a simple harmonic oscillator is given by

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

and, because $f=1/T$ , the frequency of a simple harmonic oscillator is

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}.$

Note that neither $T$ nor $f$ has any dependence on amplitude.

## Take-home experiment: mass and ruler oscillations

Find two identical wooden or plastic rulers. Tape one end of each ruler firmly to the edge of a table so that the length of each ruler that protrudes from the table is the same. On the free end of one ruler tape a heavy object such as a few large coins. Pluck the ends of the rulers at the same time and observe which one undergoes more cycles in a time period, and measure the period of oscillation of each of the rulers.

## Calculate the frequency and period of oscillations: bad shock absorbers in a car

If the shock absorbers in a car go bad, then the car will oscillate at the least provocation, such as when going over bumps in the road and after stopping (See [link] ). Calculate the frequency and period of these oscillations for such a car if the car’s mass (including its load) is 900 kg and the force constant ( $k$ ) of the suspension system is $6\text{.}\text{53}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N/m}$ .

Strategy

The frequency of the car’s oscillations will be that of a simple harmonic oscillator as given in the equation $f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$ . The mass and the force constant are both given.

Solution

1. Enter the known values of k and m :
$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{6\text{.}\text{53}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N/m}}{\text{900}\phantom{\rule{0.25em}{0ex}}\text{kg}}}.$
2. Calculate the frequency:
$\frac{1}{2\pi }\sqrt{\text{72.}6/{\text{s}}^{–2}}=1\text{.}{\text{3656}/\text{s}}^{\text{–1}}\approx 1\text{.}{\text{36}/\text{s}}^{\text{–1}}=\text{1.36 Hz}.$
3. You could use $T=2\pi \sqrt{\frac{m}{k}}$ to calculate the period, but it is simpler to use the relationship $T=1/f$ and substitute the value just found for $f$ :
$T=\frac{1}{f}=\frac{1}{1\text{.}\text{356}\phantom{\rule{0.25em}{0ex}}\text{Hz}}=0\text{.}\text{738}\phantom{\rule{0.25em}{0ex}}\text{s}.$

Discussion

The values of $T$ and $f$ both seem about right for a bouncing car. You can observe these oscillations if you push down hard on the end of a car and let go.

If a time-exposure photograph of the bouncing car were taken as it drove by, the headlight would make a wavelike streak, as shown in [link] . Similarly, [link] shows an object bouncing on a spring as it leaves a wavelike "trace of its position on a moving strip of paper. Both waves are sine functions. All simple harmonic motion is intimately related to sine and cosine waves.

The displacement as a function of time t in any simple harmonic motion—that is, one in which the net restoring force can be described by Hooke’s law, is given by

$x\left(t\right)=X\phantom{\rule{0.25em}{0ex}}\text{cos}\frac{2\mathrm{\pi t}}{T},$

where $X$ is amplitude. At $t=0$ , the initial position is ${x}_{0}=X$ , and the displacement oscillates back and forth with a period $T$ . (When $t=T$ , we get $x=X$ again because $\text{cos}\phantom{\rule{0.25em}{0ex}}2\pi =1$ .). Furthermore, from this expression for $x$ , the velocity $v$ as a function of time is given by:

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