Angular Velocity is the rate of change of angle with respect to time. The symbol for angular velocity is (pronounced “omega”).

Mathematically = t

Angular Velocity is measured in radians per second (rad/s), where 2π radians corresponds to a full circle (3600).


If the object is carrying out full revolutions, then the time corresponding to one full revolution is called the periodic time (symbol T).

Therefore in this case the expression = t becomes = 2πT

                 = 2πT               or                  T=

If an object is moving in a circle at constant speed, it is accelerating. This isbecause while its speed is not changing, its velocity is. Why? Because velocity is defined as speed in a given direction, so if direction is changing, even though speed is not, then the velocity is changing, therefore the object is accelerating.

Now for an object undergoing complete revolutions the frequency (f) of oscillation corresponds to the number of revolutions per second, where f = 1/T

e.g. if the periodic time is 0.2 seconds then the frequency will be 5 (Hertz).

Relationship between Linear Speed (v) and Angular Velocity ()

Using the relationship velocity = distance travelled/time taken (and for a full circle the distance travelled is the circumference of a circle = 2πr)

  • v = 2πr/T but 2π/T =
  • v = r    = r


Centripetal Force:The force – acting in towards the centre – required to keep an object moving in a circle is called Centripetal Force.      F=mr2

                   =mv2 /r

Centripetal Acceleration:If a body is moving in a circle then its acceleration towards the centre is called Centripetal Acceleration.



Basically what’s going on in Simple Harmonic Motion is that an object has its natural resting point, and if, when it gets disturbed from this point it tries to return but actually overshoots the mark and has to return again and again.

In our study of Newton’s Laws of Motion we consider the motion of a body when subject to a constant force or forces. As a result we can calculate the object’s velocity or position at any time.

However there are many instances when a moving object is subject to a changing force – can we still calculate future position and velocity? Well, we can if we can quantify the force, i.e. if we know how the force is changing. One example of an object experiencing a changing force is a stretched spring. We say that the spring is undergoing Simple Harmonic Motion (SHM).

An object is said to be moving with Simple Harmonic Motion if;

  1. its acceleration is directly proportional to its distance from a fixed point in its path, and
  2. its acceleration is directed towards that point.

We represent this mathematically as :       a = –2 x        where  in this case is the elastic constant (see note below)

  • The ‘fixed point’ referred to in the definition is also called the ‘equilibrium position’.

  • The key to SHM is that the acceleration and the displacement are always opposite in direction.
  • This is represented by the minus sign in the equation, although mathematically the minus sign has no significance and can be ignored for calculations.


A = Amplitude

x = distance from equilibrium position

= constant       Note:ω is the same symbol as was used previously (Circular Motion) to represent angular velocity, however it does not represent angular     velocity here. Here it is the elastic constant, (also called the constant of elasticity or rate of spring constant)

v =- velocity at any time t

T = periodic time (time for one complete oscillation)

Know the following equations

  1. A = – 2x   = distance from the Release Point to the Equilibrium Position

  2. v2 = 2(A2 – x2)
  3. x = A Sin t (if the particle starts at the centre)
  4. x = A Cos t (if the particle starts at the extreme)
  5. x = A Sin (t + ) (if the particle starts somewhere else)
  6. vmax = A
  7. amax = 2A
  8. T = 2π/

Hooke’s law of elasticity is an approximation that states that the extension of a spring is in direct proportion with the load applied to it. Many materials obey this law as long as the load does not exceed the material’s elastic limit. Materials for which Hooke’s law is a useful approximation are known as linear-elastic or “Hookean” materials. Hookean materials is a necessarily broad term that may include the work of muscular layers of the heart. Hooke’s law in simple terms says that stress is [[directly proportional] to strain]]. Mathematically, Hooke’s law states that



x is the displacement of the spring’s end from its equilibrium position (a distance, in SI units: metres);
F is the restoring force exerted by the spring on that end (in SI units: N or kg·m/s2); and
k is a constant called the rate or spring constant (in SI units: N/m or kg/s2).

When this holds, the behaviour is said to be linear. If shown on a graph, the line should show a direct variation. There is a negative sign on the right hand side of the equation because the restoring force always acts in the opposite direction of the displacement (for example, when a spring is stretched to the left, it pulls back to the right).

Make sure that your calculator is in radian mode when using these equations.


1. Find the position of equilibrium

Forces left = Forces right, or Forces up = Forces Down.

This gives a value for d, and hence a position of equilibrium.

If however, you are dealing with a string which lies horizontally on a table, then simply take the equilibrium position point to be at the end of the natural length of the string.

If we refer to this extension as ‘d’ it avoids confusion with ‘x’ which we should be using in the next step.

2. Show that the object exhibits SHM (and finding )

Extend the object a distance x from equilibrium.

Now let Fsmall – Fbig = ma

The forces are either k(ext), or else mg.

The formula is rearranged to give a = – (constant) x.

This constant now equals 2.

3. Calculate the period T.

T = 2/

4. Note the Amplitude

At this stage we look again at the question to see where the object is released from.

The Amplitude A corresponds to the distance between this release point and the equilibrium position.

5. Answer the question

Okay, it seems like a silly thing to say, but really it is only at this stage that you need consider what the question is asking you to do.

Usually we need to find the time which it takes the object to go from its release point to some final position.

Usually this involves passing through the equilibrium position and out the other side.

We deal with this in two stages:

The time to go from the release point to the equilibrium position corresponds to the time to go through one amplitude.

As this is one quarter of the total oscillation, the time for this stage corresponds to T/4


  • The time to go through one complete oscillation is the period T.

  • The time to go from one extreme to the other corresponds to the time to go through half an oscillation, and this equals T/2.

  • The time to go form the release point to the equilibrium position corresponds to the time to go through one amplitude, and this equals T/4.

However at any other sub-intervals the distance travelled is not directly proportional to the time, and this is why we revert to the equation: x = A Sint etc.

Also with these equations, x always corresponds to the distance between the given position and the equilibrium position. This can cause confusion if the object is released from an extreme and the formula: x =A Cost is used.

Know the significance of the phrase ‘when the string is taut’, and ‘when the string is slack’ (when the string is taut the object will be undergoing SHM, but not when it is slack).

Quite often the object will only be exhibiting SHM for part of its motion; you must be able to figure out when it is and when it is not exhibiting SHM. It will only be exhibiting SHM when the string is stretched.

You should be able to use VUAST for the other situations.

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