Simple Harmonic Motion

From ZuluNotes - Free Leaving Cert Notes

Simple Harmonic Motion

Subject:	Applied Mathematics & Physics
Level	Honours only
Note	Physics course covers only basic applications of Hooke's Laws

[edit] Definition

A particle is said to move with simple harmonic motion, if its acceleration is proportional to its displacement from a fixed point, and is directed towards that fixed point.

There are many examples of this type of motion in everyday life, for example the motion of the pendulum in an old clock. One can observe that as the pendulum moves to its furthest point on either side, it is slowing down in terms of speed. Eventually, it stops and then returns in the other direction back towards the center of the arc of motion, and then repeats this process again. This is an example of Simple Harmonic Motion, since its acceleration is proportional to its displacement from the centre of the arc of motion - :compare the speed of the pendulum at the furthest point on either side, to its speed at the center of motion.

Another example of Simple Harmonic Motion, is the motion of a body sitting on the surface of water, when a force is applied vertically downwards on it and then released. Using Archimedes Principle, we know that the upthrust acting on the body is proportional to how much of the water is displaced. Therefore the upthrust is varying as the body moves up and down in the water and this is an example of Simple Harmonic Motion since its acceleration is proportional to its vertical displacement from the level of equilibrium.

There are a number of formulae that apply to this topic, however all but one are derived from Hooke's Law.

[edit] Applications and Formulae

A " $\rightarrow$ " denotes formulae that must be memorised by a student.

Hooke's Law is the mathematical representation of the definition given above. In terms of the Physics course, it is defined as:

$F =\ -kx$

Where:

$F$ is the force.

$k$ is the so-called spring constant.

$x$ is the displacement from the fixed point.

In terms of the Applied Mathematics course, it is rewritten as:

$F =\ k(l-l_{0})$

Where:

$F$ is the force.

$k$ is the so-called spring constant.

$l$ is the length of the string (be it under tension or not).

$l_{0}$ is the natural length of the string.

The spring constant is a constant that is particular to a certain spring. It is the ratio of the force exerted per distance extended. Its unit of measurement is the Newton per unit metre.

We can derive three further equations from the above:

$F =\ -kx$

$ma =\ -kx$

$a =\ -\frac{k}{m}.x$

$\rightarrow\ a =\ -\omega^{2}x$

$\rightarrow\ \omega\ =\ \sqrt{\frac{k}{m}}$

We let the fraction part of the formula be equal to $\omega^{2}$ so that to ensure that the acceleration is always positive when the displacement is negative and vice versa.

Students should realise that it is very important to be aware what omega is equal to, because of the fact that its value is required to be able to determine the period of motion:

$\rightarrow\ T\ =\ \frac{2\pi}{\omega}$

Below here applies to the Applied Mathematics course only.

We can write the acceleration as the following derivative: (for more on this see the page on ODE's)

$v\frac{dv}{dx}\ =\ -\omega^{2}x$

$\int\ v\ dv\ =\int\ -\omega^{2}x\ dx$

$\frac{v^2}{2}\ =\ \frac{-\omega^{2}x}{2}\ +\ C$

We know from observation that the velocity of an object under motion that is simple harmonic, is zero when it is displaced a maximum distance from the point of equilibrium, that is, when its displacement is equal to its amplitude. We can therefore determine the constant of integration. We then have the following equation.

$\frac{v^2}{2}\ =\ \frac{-\omega^2 x^2}{2}\ +\ \frac{\omega^2 A^2}{2}$

$v^2\ =\ \omega^2 A^2\ -\ \omega^2 x^2$

$\rightarrow\ v\ =\ \omega \sqrt{A^2\ -\ x^2}$

Velocity can be written as the rate of change of distance with time.

$\frac{dx}{dt} =\ \omega \sqrt{A^2\ -\ x^2}$

$\int \frac{dx}{\sqrt{A^2\ -\ x^2}} =\int \omega\ dt$

$Sin^{-1}(\frac{x}{A})\ =\ \omega t\ +\ C$

$\frac{x}{A}\ =\ Sin(\omega t\ +\ C)$

$\rightarrow\ x\ =\ ASin(\omega t\ +\ \delta)$

The value of the constant of integration depends on the initial conditions of the object under simple harmonic motion, and it is commonly denoted by the Greek letter delta.

If we rewrite the above in terms of Cosine, it is possible to use this new formula to find the distance the object is from the furthest point of displacement:

$\rightarrow\ x\ =\ ACos(\omega t\ +\ \delta)$

Students should be able to determine when the use of one equation over the other would be more beneficial in terms of solving a problem.

Very generally, if the object's initial position is at its point of maximum displacement, then the Cosine version should be used, and if its initial position is at the point of equilibrium, then the Sine version should be used. The reason for this is because, when the object's initial position is at its point of maximum displacement, using the Cosine formula means that the value of $\delta$ is zero, and likewise the Sine formula when the object's initial position is at the point of equilibrium. When the objects initial position is somewhere in between, then the conditions/requirements of the question will dictate which version should be used.

Manipulating the above and realising that the greatest velocity is when the object is at the point of equilibrium and the greatest acceleration is at the greatest displacement we have the two additional equations:

$\rightarrow\ v_{max}\ =\ \omega A$

$\rightarrow\ a_{max}\ =\ -\omega^2 A$

[edit] Equations Checklist

Equation	Physics	Applied Mathematics
$F =\ -kx$
$F =\ k(l-l_{0})$
$a =\ -\omega^{2}x$
$\omega\ =\ \sqrt{\frac{k}{m}}$
$T\ =\ \frac{2\pi}{\omega}$
$v\ =\ \omega \sqrt{A^2\ -\ x^2}$
$x\ =\ ASin(\omega t\ +\ \delta)$
$v_{max}\ =\ \omega A$
$a_{max}\ =\ -\omega^2 A$

[edit] Exam Paper

[edit] Physics

There is no question in the Physics exam paper in which Simple Harmonic Motion is examined exclusively. However, it is most often

[edit] Applied Mathematics

Question 6 on the honours Applied Mathematics paper is the Simple Harmonic Motion and Circular Motion question. This question can frequently be quite difficult, and it is therefore not one of the most common topics or questions taken by students. To do well in this question, one must have a very good understanding of the theory behind the concepts and an understanding of where the equations are coming from.

Question 6 consists almost always of two parts: a part A and a part B. In part A of the question, one topic that is examined quite frequently is to show that a body is moving with Simple Harmonic Motion. This is very straightforward and can be shown quite easily in most cases. Part B of the question usually deals with Simple Harmonic Motion, with Circular Motion usually being found in the part A, if at all. On a number of occasions on past papers, some very difficult questions have been examined in part B. The following worked example is taken from the 1996 exam paper:

6. (a) A body of mass 10kg moves with Simple Harmonic Motion. At a displacement of 0.8m from the centre
       of oscillation, the velocity and acceleration of the body are 2 m/s and 20 m/s^(2) respectively.
          Find
            (i)  the number of oscillations per second
           (ii)  the amplitude of motion
          (iii)  the maximum acceleration and hence show that the force to overcome the inertia of the
                 body at the extremity of the oscillation is 223.6N.

Solution: Part (i)

$a =\ -\omega^{2}x$

We can disregard the negative since the direction is meaningless in this question.

$20\ =\ \omega^{2}(0.8)$

$25\ =\ \omega^{2}$

$\omega\ =\ 5$

$T\ =\ \frac{2\pi}{\omega}\ =\ \frac{2\pi}{5}$

Frequency is by definition simply the inverse of the period.

$f\ =\ \frac{1}{T}\ =\ \frac{5}{2\pi}\ = 0.7958\ Hz$

Solution: Part (ii)

$v\ =\ \omega \sqrt{A^2\ -\ x^2}$

$2\ =\ 5 \sqrt{A^2\ -\ (0.8)^2}$

$4\ =\ 25 (A^2\ -\ 0.64)$

$\frac{4}{25}\ +\ 0.64\ =\ A^2$

$A^2\ =\ 0.8$

$A\ =\ \frac{2}{\sqrt{5}}\ m$

Solution: Part (iii)

$a_{max}\ =\ -\omega^2 A$

$a_{max}\ =\ (5)^2.(\frac{2}{\sqrt{5}})\ =\ 10\sqrt{5}\ m/s^{2}$

$F\ =\ ma\ =\ (10).(10\sqrt{5})\ =\ 223.6\ N$

6. (b) A light perfectly elastic string of natural length a and elastic constant k is fastened
       at one end p to a fixed point of a smooth horizontal table, and a particle of mass m is attached
       to the other end. The particle is held on the table at a distance 2a from p and then released.
         Prove
           (i)  that the particle executes simple harmonic motion while the string is taut
          (ii)  that the particle reaches p after  seconds.

Solution: Part (i)

The force exerted on the particle is:

$F\ =\ -kx$

by Hooke's Law. Therefore:

$ma\ =\ -kx$

$a\ =\ -\frac{kx}{m}$

This equation shows us that the acceleration of the body is proportional to its displacement from the origin and directed towards the origin. This is consistent with Simple Harmonic Motion.

Solution: Part (ii)

This seems like a difficult question because one must work entirely with variables, however if you do not lose yourself in equations, it is quite straightforward.

We know that:

$\omega\ =\ \sqrt{\frac{k}{m}}$

Therefore the period is:

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

A period is the length of time it would take the particle to move from its initial position through the origin and then back to its initial position. Therefore, the time it takes the particle to move from its initial position to its first meeting with the point of equilibrium is only one-fourth of the period:

$\frac{T}{4}\ =\ \frac{\pi}{2}\sqrt{\frac{m}{k}}$

Be aware that when the particle moves past the point of equilibrium, the string will go slack, and therefore no longer exert a force on the particle. We now must consider its speed at that point. We note also that the question says that we are dealing with a smooth table, and therefore there is no friction.

The velocity at the point that the string goes slack:

$v_{max}\ =\ \omega A$

$v_{max}\ =\ A\sqrt{\frac{k}{m}}$