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 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.

pendulum1

pendulum

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 vmax 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

fig4

By Pythagoras Theorem (refer to figure 3)

figure5

fig6

fig7

fig8

Byy definition the time taken = distance covered /speed

So for one cycle (T) = circumference of circle/speed

fig9 ----- Eq1

Now recall that vmax = √(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

 

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