Unit 1.1 SHM & the Simple Pendulum
Task:
Determine (a) the parameters of the frequency of the simple pendulum (b) whether or not the motion is simple harmonic. State any assumptions you make.
Assumptions:
The cord is massless. The cord is rigidly attached to the support at A. The cord does not stretch.
Consider the pendulum bob is displaced from its equilibrium position D to a new position C.
Figure 1

Figure 2

Length of cord = l
Angle of displacement = θ.
Mass of bob = m
Arc of displacement = s
Horizontal displacement = x
The tension on the string is FT
Weight of bob = mg
The weight of the bob can be resolved into two components mgCosθ (along the length of the cord) and mgsinθ (perpendicular to the cord and actually tangential to the arc).
By Newton’s First law
1. The string is inextensible so FT – mgCosθ = 0 as it does not move in the direction of the string but we are not interested in this force!
2. Perpendicular to the direction of the string we have an unbalanced force of mgsinθ towards the central position so the restoring force Fr = mgsinθ (The minus sign comes about because θ was measured in the anticlockwise direction from the equilibrium and force Fr points in the opposite clockwise direction).
The displacement s = l θ when θ is measured in radians
Eq1
As explained above, Fr = –mg sin θ Eq2
For angles (less than 15° when θ is measured in radians) θ ≈ Sin θ. You can see this easily in the diagram where it is quite easy to imagine the length of the arc s is approximately equal to the displacement x (side BC of the triangle ABC. From trigonometry of the triangle ABC, x = l Sinθ).
Fr ≈ mgθ and l Sinθ ≈ = l θ
So proceeding on the assumption that θ is less than 15^{o}.
Eq3
Substituting for θ from Eq1 into Eq3 we get
Rewriting
Since the displacement s ≈ x for small angles, we can rewrite for this system as
Note m, g and l are all constants in this investigation so the equation is of the form F = kx
And recall in Hooke’s Law the restoring force Fr = kx
So
Recall that for simple harmonic motion (* I will prove this in another document  Derivation of Period and Frequency for SHM  Eq2)
Substituting for k we get .
Eq4
Recall that f = ^{1}/_{T}
Eq5
Conclusions:
(a) By definition if the restoring force is proportional to s or to θ the motion will be simple harmonic. We found out that so the restoring force is proportional to Sinθ from the equation Fr = mgSinθ. It is only for small angles that Fr = mgθ. So we can only conclude that the motion approximated SHM for small angles (>15°)
(b) The mass of the bob does not appear in equations 3 & 4 so we can conclude that the period and frequency of the simple pendulum are independent of the mass. The period and frequency depend only on the length of the chord and acceleration due to gravity.
Exercise 1.1.12
