Displacement in Simple Harmonic Motion

The behaviour of a simple harmonic oscillator is expressed in terms of its displacement $x$ from equilibrium, its velocity $x^{},$ and its acceleration $x^{}$ at any given time. If we try the solution

$x=$Acos$t$

where $A$ is a constant with the same dimensions as $x,$ we shall find that it satisfies the equation of motion

$x^{}+{}^{2}x=0$

for

$x^{}=-Asint$

and

$x^{}=-A{}^{2}cost=-{}^{2}x$

Displacement in Simple Harmonic Motion

$5$

Another solution

$x=$Bsin$t$

is equally valid, where $B$ has the same dimensions as $A,$ for then

$x^{}=Bcost$

and

$x^{}=-B{}^{2}sint=-{}^{2}x$

The complete or general solution of equation $(1\mathrm{.}1)$ is given by the addition or superposition of both values for $x$ so we have

$x=$Acos$t+$Bsin$t$

with

$x^{}=-{}^2(Acost+$Bsin$t)=-{}^{2}x$

where A and $B$ are determined by the values of $x$ and $x^{}$ at a specified time. If we rewrite the constants as

$A=$asin$,$ and $,B=$acos$$

where $$ is a constant angle, then

$A^2+B^2=a^2(si{n}^2+co{s}^2)=a^2$

so that

$a=\sqrt[\phantom{2}]{A^2+B^2}$

and

$x=$asin$cost+$acos$sint$

$=$asin$(t+)$

The maximum value of $sin(t+)$ is unity so the constant $a$ is the maximum value of $x,$ known as the amplitude of displacement. The limiting values of $sin(t+)$ are $\backslash$pm $1$ so the system will oscillate between the values of $x=+-$a and we shall see that the magnitude of $a$ is determined by the total energy of the oscillator.

The angle $$ is called the 'phase constant' for the following reason. Simple harmonic motion is often introduced by reference to 'circular motion' because each possible value of the displacement $x$ can be represented by the projection of a radius vector of constant length $a$ on the diameter of the circle traced by the tip of the vector as it rotates in a positive

Figure $1\mathrm{.}2$ Sinusoidal displacement of simple harmonic oscillator with time, showing variation of starting point in cycle in terms of phase angle $$

anticlockwise direction with a constant angular velocity $.$ Each rotation, as the radius vector sweeps through a phase angle of $2rad,$ therefore corresponds to a complete vibration of the oscillator. In the solution

$x=$asin$(t+)$

the phase constant $,$ measured in radians, defines the position in the cycle of oscillation at the time $t=0,$ so that the position in the cycle from which the oscillator started to move is

$x=$asin$$

The solution

$x=$asin$t$

defines the displacement only of that system which starts from the origin $x=0$ at time $t=0$ but the inclusion of $$ in the solution

$x=$asin$(t+)$

where $$ may take all values between zero and $2$ allows the motion to be defined from any starting point in the cycle. This is illustrated in Figure $1\mathrm{.}2$ for various values of $.$

Can you simplify and drive these equation step by step