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 Energy in Simple Harmonic Motion


(a) Determine velocity of the object at any position x when an object is in simple harmonic motion. (b) Explain the relationship between amplitude and energy possessed by the object.


There is no friction in the system. The motion is sinusoidal and of a single frequency.

Consider a spring mass system in simple harmonic motion. As the mass oscillates back and forth the energy is continually interchanged between potential (Ep) and kinetic (Ek) energies. According to the Law of conservation of enery the total energy is constant as there is no friction or other losses in this system.

The object stops for an instant at the extremes where velocity = 0 and has maximum velocity vmax as it passes o (in either direction).

When the object is at rest the Ek = 0 and Ep = max

When the object is at maximum velocity Ep = 0 and Ek = max


Ep = Work done

Work done = Average Force.displacement

Ep = ((F + 0)/2)x = ½Fx

Substuting F = kx from Hooke's Law

Ep = (kx)x

Ep = ½kx2

and Ek = ½mv2


Now By the law of conservation of energy Et = Ep + Ek

At -A, Et = Ep + Ek

Et = ½kA2 + 0

At -A, Et = ½kA2 --- Eq1

At O,  Et = 0 + ½mv2max

At OEt = ½mv2max ----Eq2

 Combining Eq1 and Eq2 we get ½kA2 = ½mv2max

=> v2max = (k/m)A2 ----Eq3

vmax = √(k/m)A ---- Eq4

At A, Et = ½kA2 + 0

At x, Et = ½kx2 + ½mv2

½kA2 = ½kx2 + ½mv2


fig3   ----- Eq5

Substituting Eq3 into Eq5 we get   fig4

fig5    ----- Eq6


(a) From Eq6 we see that at any position x, the object moves back and forth, so its velocity can be either in the + or - direction, but its magnitude depends only on its position x.

Vmax when x = A or –A and v = 0 when x = 0.

(b) In Eq1 we see that Et = ½kA2 so clearly the total energy is proportional to the amplitude of the wave.


Exercise 1.1.11

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