1.1 Introduction
1.1 Oscillations or Vibrations
1.1 Oscillations or Vibrations some important points
1.1 Wave Quantities
1.1 Wave Properties
1.1 Wave Types
1.1 Speed of Waves
1.1 Simple Harmonic Motion
1.1 Derivation of Period and Frequency for SHM
1.1 Energy in Simple Harmonic Motion
1.1 SHM & Simple Pendulum
1.1 Reflection and Transmission
1.1 Summary
Unit 1.1 Multiple Choice Questions - Waves
Unit 1.1 Multiple Choice Answers - Waves

 

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.

fig1

Figure 1

fig2

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

fig3-------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 15o.

fig 3 -------Eq3

Substituting for θ from Eq1 into Eq3 we get fig5

Rewriting fig6

Since the displacement s ≈ x for small angles, we can rewrite for this system as fig6

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 fig8

Recall that for simple harmonic motion fig9 (* I will prove this in another document - Derivation of Period and Frequency for SHM - Eq2)

Substituting for k we get fig10.

fig11-----Eq4

Recall that f = 1/T

fig12-----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

 

Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.