Unit 1.1 Derivation of Period and Frequency for SHM
Task:
Derive a formula for the period of simple harmonic motion (SHM) by comparing SHM to an object rotating uniformly in a circle. State any assumptions you make.
Assumptions:
The object is traveling at constant speed in a circle in a horizontal plane.
Consider a small object on a string moving in circular anticlockwise motion in a horizontal plane just above your table surface covered in graph paper at constant speed. From the top the object appears traverse the circular path in the xy plane over the graph illustrated in figure 1 with a velocity vmax. However, from the side it will appear to move in a line along the x axis as in figure 2 with a velocity v. Consider the object starting at O which is directly in front of you. It appears to move from O —› B —› O —› C —› O.

Figure 3

Let us examine the object when it is at an angle θ. The observer will see it at A on the x axis
The velocity of the particle is v_{max} at a tangent to the circular path.
The velocity along the x axis is v.
Examine the two triangles OAC and CDE.
Both are similar (right angled and with one angle θ).
=> ^{CE}/_{CD} = ^{AC}/_{OC}
By Pythagoras Theorem (refer to figure 3)
Byy definition the time taken =^{ distance covered }/_{speed}
So for one cycle (T) = ^{circumference of circle}/_{speed}
 Eq1
Now recall that v_{max} = √(^{k}/_{m})A (* I will prove this in another document  Energy in Simple Harmonic Motion Eq4)
In this example the amplitude is r so A becomes r in the equation.
=> ^{Vmax}/_{A} = √(^{k}/_{m})
=> ^{A}/_{Vmax} = √(^{m}/_{k})
Now A the amplitude is really the radius of the circle r.
=> ^{r}/_{Vmax} = √(^{m}/_{k})
Substituting for ^{r}/_{Vmax} Eq1 becomes T = 2Π√(^{m}/_{k})  Eq2
Conclusion: The period depends on the mass of the mass of the spring and its spring constant k but is independent of the amplitude.
Exercise 1.1.10
