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Wize University Physics Textbook (Master) > Periodic Motion: Oscillations

Simple Harmonic Motion for Small Deviations from Equilibrium

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Simple Harmonic Motion for Small Deviations from Equilibrium


There are many physical systems in which we observe some type of oscillation. These oscillation could be described as a simple harmonic motion if the displacement or deviation from equilibrium is very small.

Example:

























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Finding the Angular Frequency of Oscillation

These are the steps to find the angular frequency of these oscillation if we know the general form of the force (F(z)F(z)) exerted on the oscillator:

  1. Find equilibrium position (z0z_0) by setting the force equal to zero:
F(z0)=0F(z_0)=0

  1. Write down the Taylor polynomial of the force function about the equilibrium point:
F(z)=F(z0)+F′(z0)1!(z−z0)+F′′(z0)2!(z−z0)2+...+F(n)(z0)n!(z−z0)nF(z)=F\left(z_0\right)+\frac{F'\left(z_0\right)}{1!}\left(z-z_0\right)+\frac{F ''\left(z_0\right)}{2!}\left(z-z_0\right)^2+...+\frac{F^{\left(n\right)}\left(z_0\right)}{n!}\left(z-z_0\right)^n


  1. (z−z0)(z-z_0)is displacement from equilibrium. Let me show this displacement by a new variable xx. If we assume this displacement is very small, its higher powers is getting smaller and smaller and as a result, the dominant term in the above expansion is the second term (The first term is zero using equilibrium equation).
For small displacements:      F(x)=F′(z0)1!(z−z0)=F′(z0)x\text{For small displacements:}\ \ \ \ \ \ F(x)= \frac{F'\left(z_0\right)}{1!}\left(z-z_0\right)=F'\left(z_0\right)x


  1. By comparing above force expression by the general form of SHM force (F=−kxF=-kx), we can find the effective spring constant of this oscillation as:

keff=−F′(z0)\boxed{k_{eff}=-F'(z_0)}

  1. Finally the angular frequency of this oscillation is:

ω=keffm=−F′(z0)m\boxed{\omega=\sqrt{\dfrac{k_{eff}}{m}}=\sqrt{\dfrac{-F'(z_0)}{m}}}


Watch Out!
You don't to show all these steps in exam. As long as you remember that keff=−F′(z0)k_{eff}=-F'(z_0)you should be able to easily find the angular frequency of the oscillation.

Exam Tip
We use this method for problems in which the force acting on the oscillator does not have the typical form of a SHM force:
F=−kxF=-kx



Example: Small Oscillations


Consider a 1010 kg piston pressing down on a cylinder filled with water, initially at equilibrium at height h0h_0. As the piston is slightly nudged away from its equilibrium position, it will oscillate in SHM, with a position-dependent force given by F(h)=ch−1000F(h)=\dfrac{c}{h}-1000. Here hh is the height from the equilibrium, and cc is a constant.

a) Find the constant cc if h0=1h_0=1 m.

b) The piston is pushed down by a distance of 11 cm and released. Find the period of oscillation as an exact value.



Part a)


As always for SHM, the force should be zero at equilibrium, so F(h0)=0F(h_0)=0 with h0=1h_0=1.

F(h)=ch−1000F(h)=\dfrac{c}{h}-1000

0=c1−10000=\dfrac{c}{1}-1000

c=1000c=1000 (Nm)

Therefore the force is: F(h)=1000h−1000F(h)=\dfrac{1000}{h}-1000

Part b)


We have h−h0=0.01h-h_0=0.01 m which is much smaller than the equilibrium height h0=1h_0=1 m. Therefore we can use Taylor expansion about h0\bcf {h_0} to find the approximate spring constant keff\bcf {k_{eff}} for the system.

Differentiate the force:

F′(h)=−1000h2F'(h)=-\dfrac{1000}{h^2}

Now we can find keffk_{eff} as:

keff=−F′(h0)k_{eff}=-F'(h_0)

=1000(1)2=\dfrac{1000}{(1)^2}

=1000=1000 (N/m)


The angular frequency is given by ω=keffm\omega=\sqrt{\dfrac{k_{eff}}{m}} , therefore the period will be:

T=2πωT=\dfrac{2\pi}{\omega}

=2πmkeff=2\pi\sqrt{\dfrac{m}{k_{eff}}}
=2π101000=2\pi\sqrt{\dfrac{10}{1000}}

=2π10=\dfrac{2\pi}{10}

=π5=\dfrac{\pi}{5} (s)