Harmonic motion occurs when a particle oscillates periodically.
Frequency \(f\) is the number of oscillations per second (Hz).
The SI unit for frequency is the hertz (Hz), which is 1 oscillation per second.
The period \(T\) is:
\[
T = \frac{1}{f}
\]
Simple harmonic motion occurs when the magnitude of the acceleration of the particle is proportional to the distance from the fixed point and acts in the opposite direction to the displacement. Simple harmonic motion is sinusoidal.
As the particle moves from P to O and back again, its motion in the x direction can be graphed as a function of time :
Speed is the magnitude of velocity, which is the rate of change of displacement with time.
The particle's speed is greatest at the midpoint and zero at points O and P.
Acceleration is shown as the blue line.
The displacement is a cosine graph.
Displacement as a Cosine Function
\[
x = \cos(nt)
\]
\[
\dot{x} = -n\sin(nt)
\]
\[
\ddot{x} = -n^2\cos(nt)
\]
\[
\ddot{x} = -n^2 x
\]
The following represents the displacement of the particle from the midpoint in the x direction as it bounces between the limits O and P. Timing is started at the midpoint, with the particle headed towards P.
The displacement is a sine graph.
Displacement as a Sine Function
\[
x = \sin(nt)
\]
\[
\dot{x} = n\cos(nt)
\]
\[
\ddot{x} = -n^2\sin(nt)
\]
\[
\ddot{x} = -n^2 x
\]
In either case,
General SHM Equation
\[
a = -n^2 x
\]
This gives the differential equation:
\[
\frac{d^2x}{dt^2} = -n^2 x
\]
\[
\frac{d^{2}x}{dt^{2}} + n^{2}x = 0
\]
where n is the number of cycles in the period.
Angular Displacement and Angular Frequency
As the particle moves, the angle moves.
\[ \theta = \frac{\text{arc length}}{radius} \]
The difference between the old and new angle is called angular
displacement:
\[ \Delta \theta = \theta_2 -\theta_1 \]
The rate of change of angular displacement is called angular frequency.
\[
\frac{d \theta}{dt} = \omega
\]
Since simple harmonic motion is sinusoidal,
\[
y(t) = \sin(\theta(t))
\]
\[
= \sin(\,wt\,)
\]
The period is
\[
T = \frac{2\pi}{n}
\]
\[
n = \frac{2\pi}{T}
\]
This can be written as
\[
T = \frac{2\pi}{\omega}
\]
\[
\omega = \frac{2\pi}{T}
\]
and substituted into
giving
\[
\frac{1}{f} = \frac{2\pi}{\omega}
\]
\[
\omega = 2\pi f
\]
Working backwards
\[
\ddot{x} = -n^{2}x
\]
\[
\ddot{x} = v\,\frac{dv}{dx}
\]
\[
v\,\frac{dv}{dx} = -n^{2}x
\]
so
\[
v\,\frac{dv}{dx} = -n^{2}x
\]
\[
\int v\,dv = \int -n^{2}x\,dx
\]
\[
\frac{v^{2}}{2} = \frac{-n^{2}x^{2}}{2} + c
\]
Further, when v = 0, the particle is at a maximum or minimum. The displacement is equal to the amplitude, a.
\[
\text{when } x = \pm a \quad v = 0
\]
\[
0 = \frac{-n^{2}a^{2}}{2} + c
\]
\[
c = \frac{n^{2}a^{2}}{2}
\]
\[
\frac{v^{2}}{2}
= \frac{-n^{2}x^{2}}{2}
+ \frac{n^{2}a^{2}}{2}
\]
\[
v^{2}
= -n^{2}x^{2}
+ n^{2}a^{2}
\]
\[
v^{2}
= n^{2}(a^{2} - x^{2})
\]
Replacing n with ω
\[
v^{2} = \omega^{2}(a^{2} - x^{2})
\]
Since the maximum speed occurs when the particle is at the midpoint,
\[
v^{2} = n^{2}(a^{2}
\]
\[
v = n a
\]
or
\[
v^{2} = \omega^{2}(a^{2}
\]
\[
v = \omega a
\]
Maximum Speed
\[
v_{\max} = a\omega
\]
also
\[
v^{2} = n^{2}(a^{2} - x^{2})
\]
\[
v = \pm\sqrt{\,n^{2}(a^{2} - x^{2})\,}
\]
\[
\frac{dx}{dt}
= \sqrt{\,n^{2}(a^{2} - x^{2})\,}
\]
\[
\frac{dx}{dt}
= \sqrt{n^{2}} \times \sqrt{\,a^{2} - x^{2}\,}
\]
\[
\int \frac{x}{\sqrt{\,a^{2} - x^{2}\,}}\,dx
= \int n\,dt
\]
\[
\sin^{-1}\!\left(\frac{x}{a}\right) = nt + c
\]
renaming the constant of integration as ε gives
\[ \sin^{-1}\!\left(\frac{x}{a}\right) = nt + \epsilon \]
Further,
\[
\sin^{-1}\!\left(\frac{x}{a}\right) = nt + \varepsilon
\]
\[
\frac{x}{a} = \sin(nt + \varepsilon)
\]
\[
x = a\,\sin(nt + \varepsilon)
\]
so
\[
x = a\sin(nt + \varepsilon)
\]
or
\[
x = a\sin(\omega t + \varepsilon)
\]
Where ε is the phase angle, a is the amplitude and n is the number of cycles in the period. (ω is the angular frequency.)
The velocity is
\[
\dot{x} = a n \cos(nt + \varepsilon)
\]
or
\[
\dot{x} = a \omega \cos(\omega t + \varepsilon)
\]
Where ε is the phase angle.
Summary
Cosine Form
\[
x = a\cos(\omega t + \varepsilon)
\]
\[
v = -a\omega\sin(\omega t + \varepsilon)
\]
\[
a = -a\omega^2\cos(\omega t + \varepsilon)
\]
Sine Form
\[
x = a\sin(\omega t + \varepsilon)
\]
\[
v = a\omega\cos(\omega t + \varepsilon)
\]
\[
a = -a\omega^2\sin(\omega t + \varepsilon)
\]
\[
F = ma
\]
\[
= m\!\left(-\omega^{2}x\right)
\]
\[
= -m\omega^{2}x
\]
Compare with Hooke’s Law:
\[
F_x = -kx
\]
The spring constant is
\[
- kx = -m\omega^{2}x
\]
\[
k = m\omega^{2}
\]
so
\[
\omega^{2} = \frac{k}{m}
\]
\[
\omega = \sqrt{\frac{k}{m}}
\]
\[
\omega = 2\pi f
\]
\[
\omega^{2} = \frac{k}{m}
\]
\[
(2\pi f)^{2} = \frac{k}{m}
\]
\[
k = 4\pi^{2} m f^{2}
\]
\[
\omega = \frac{2\pi}{T}
\]
\[
\omega^{2} = \frac{k}{m}
\]
\[
\left(\frac{2\pi}{T}\right)^{2} = \frac{k}{m}
\]
\[
4\pi^{2}m = T^{2}k
\]
\[
T = \sqrt{\frac{4\pi^{2}m}{k}}
\]
\[
T = 2\pi\sqrt{\frac{m}{k}}
\]
Example
A tub of mass 650g is attached to a spring with spring constant \(k = 65\text{ N/m}\). The tub is displaced 11cm from its equilibrium position of \( x = 0 \) and released at time \(t = 0\)
Calculate the angular frequency, frequency and period of the motion.
Angular Frequency
\[
\omega = \sqrt{\frac{k}{m}}
= \sqrt{\frac{65}{0.65}}
= 10\text{ rad/s}
\]
Frequency
\[
\omega = 2\pi f
\]
\[
f = \frac{10}{2\pi}
\]
\[
f = \frac{5}{\pi} = 1.5915
\]
\[
f = 1.6\ \text{Hz}\ (1\text{ d.p.})
\]
Period
\[
T = \frac{2\pi}{\omega}
\]
\[
= \frac{2\pi}{10}
\]
\[
= 0.628
\]
\[
= 0.6\ \text{s}\ (1\text{ d.p.})
\]
