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In fact, the mass m size 12{m} {} and the force constant k size 12{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 = m k size 12{T=2π sqrt { { {m} over {k} } } } {}

and, because f = 1 / T size 12{f=1/T} {} , the frequency of a simple harmonic oscillator is

f = 1 k m . size 12{f= { {1} over {2π} } sqrt { { {k} over {m} } } } {}

Note that neither T size 12{T} {} nor f size 12{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 size 12{k} {} ) of the suspension system is 6 . 53 × 10 4 N/m size 12{6 "." "53" times "10" rSup { size 8{4} } `"N/m"} {} .

Strategy

The frequency of the car’s oscillations will be that of a simple harmonic oscillator as given in the equation f = 1 k m size 12{f= { {1} over {2π} } sqrt { { {k} over {m} } } } {} . The mass and the force constant are both given.

Solution

  1. Enter the known values of k and m :
    f = 1 k m = 1 6 . 53 × 10 4 N/m 900 kg . size 12{f= { {1} over {2π} } sqrt { { {k} over {m} } } = { {1} over {2π} } sqrt { { {6 "." "53" times "10" rSup { size 8{4} } "N/m"} over {"900"" kg"} } } } {}
  2. Calculate the frequency:
    1 72. 6 / s –2 = 1 . 3656 / s –1 1 . 36 / s –1 = 1.36 Hz . size 12{ { {1} over {2π} } sqrt {"72" "." 6/s rSup { size 8{2} } } =1 "." "36"/s=1 "." "36 Hz"} {}
  3. You could use T = m k size 12{T=2π sqrt { { {m} over {k} } } } {} to calculate the period, but it is simpler to use the relationship T = 1 / f size 12{T=1/f} {} and substitute the value just found for f size 12{f} {} :
    T = 1 f = 1 1 . 356 Hz = 0 . 738 s . size 12{T= { {1} over {f} } = { {1} over {1 "." "36"" Hz"} } =0 "." "737"" s"} {}

Discussion

The values of T size 12{T} {} and f size 12{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 figure shows the front right side of a running car on an uneven rough surface which also shows the driver in the driving seat. There is an oscillating sine wave drawn from left to the right side horizontally throughout the figure.
The bouncing car makes a wavelike motion. If the restoring force in the suspension system can be described only by Hooke’s law, then the wave is a sine function. (The wave is the trace produced by the headlight as the car moves to the right.)
There are two iron paper roll bars standing vertically with a paper strip stitched from one bar to the other. There is a vertical hanging spring just over the middle of the two bars, perpendicular to the strip of the paper, having an object with mass m tied to it. There is a line graph with amplitude scale as X, zero and negative X on the left side of the paper strip, vertically over each other with their points marked. A perpendicular line is drawn through this amplitude scale toward the right with a point T marked over it, showing the time duration of the amplitude. This line has an oscillating wave drawn through it.
The vertical position of an object bouncing on a spring is recorded on a strip of moving paper, leaving a sine wave.

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 t = X cos 2 πt T , size 12{x left (t right )=X"cos" { {2π`t} over {T} } } {}

where X size 12{X} {} is amplitude. At t = 0 size 12{t=0} {} , the initial position is x 0 = X size 12{x rSub { size 8{0} } =X} {} , and the displacement oscillates back and forth with a period T . (When t = T , we get x = X size 12{x=X} {} again because cos = 1 .). Furthermore, from this expression for x size 12{x} {} , the velocity v size 12{v} {} as a function of time is given by:

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Source:  OpenStax, General physics ii phy2202ca. OpenStax CNX. Jul 05, 2013 Download for free at http://legacy.cnx.org/content/col11538/1.2
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