Course Content
Class 11 Physics Chapter 4 Motion In A Plane
4 Motion in a plane 4.1 Introduction 4.2 Scalars and vectors 4.3 Multiplication of vectors by real numbers 4.4 Addition and subtraction of vectors – graphical method 4.5 Resolution of vectors 4.6 Vector addition – analytical method 4.7 Motion in a plane 4.8 Motion in a plane with constant acceleration 4.9 Relative velocity in two dimensions 4.10 Projectile motion 4.11 Uniform circular motion
Class 11 Physics Chapter 5 Laws of motion
Section Name Topic Name 5 Laws of motion 5.1 Introduction 5.2 Aristotle’s fallacy 5.3 The law of inertia 5.4 Newton’s first law of motion 5.5 Newton’s second law of motion 5.6 Newton’s third law of motion 5.7 Conservation of momentum 5.8 Equilibrium of a particle 5.9 Common forces in mechanics 5.10 Circular motion 5.11 Solving problems in mechanics
Class 11 Physics Chapter 6 Work Energy and Power
Section Name Topic Name 6 Work Energy and power 6.1 Introduction 6.2 Notions of work and kinetic energy : The work-energy theorem 6.3 Work 6.4 Kinetic energy 6.5 Work done by a variable force 6.6 The work-energy theorem for a variable force 6.7 The concept of potential energy 6.8 The conservation of mechanical energy 6.9 The potential energy of a spring 6.10 Various forms of energy : the law of conservation of energy 6.11 Power 6.12 Collisions
Class 11 Physics Chapter 7 Rotation motion
Topics Introduction Centre of mass Motion of COM Linear Momentum of System of Particles Vector Product Angular velocity Torque & Angular Momentum Conservation of Angular Momentum Equilibrium of Rigid Body Centre of Gravity Moment of Inertia Theorem of perpendicular axis Theorem of parallel axis Moment of Inertia of Objects Kinematics of Rotational Motion about a Fixed Axis Dynamics of Rotational Motion about a Fixed Axis Angular Momentum In Case of Rotation about a Fixed Axis Rolling motion
Class 11 Physics Chapter 9 mechanics properties of solid
Section Name Topic Name 9 Mechanical Properties Of Solids 9.1 Introduction 9.2 Elastic behaviour of solids 9.3 Stress and strain 9.4 Hooke’s law 9.5 Stress-strain curve 9.6 Elastic moduli 9.7 Applications of elastic behaviour of materials
Class 11 Physics Chapter 11 Thermal Properties of matter
Section Name Topic Name 11 Thermal Properties of matter 11.1 Introduction 11.2 Temperature and heat 11.3 Measurement of temperature 11.4 Ideal-gas equation and absolute temperature 11.5 Thermal expansion 11.6 Specific heat capacity 11.7 Calorimetry 11.8 Change of state 11.9 Heat transfer 11.10 Newton’s law of cooling
Class 11 Physics Chapter 14 Oscillations
Section Name Topic Name 14 Oscillations 14.1 Introduction 14.2 Periodic and oscilatory motions 14.3 Simple harmonic motion 14.4 Simple harmonic motion and uniform circular motion 14.5 Velocity and acceleration in simple harmonic motion 14.6 Force law for simple harmonic motion 14.7 Energy in simple harmonic motion 14.8 Some systems executing Simple Harmonic Motion 14.9 Damped simple harmonic motion 14.10 Forced oscillations and resonance
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About Lesson

Different Expressions for Wave Speed

  • Wave speed V= ω/k. By definition of ω= (2 π)/T where T is period of the wave.
  • Also k=(2 π)/λ .Therefore v=(2 π)/(T)/(2 π)/(λ) =λ/T
  • v= λv
    • As wavelength of a wave increases as a result frequency of the wave decreases as a result speed of wave is constant.
    • Wave speed is determined by the properties of the medium.

Dimensional Analysis to show how the speed is related to mass per unit length and Tension

  • μ = [M]/[L] = [ML-1] (i)
  • T=F=ma =[M][LT-2] = [MLT-2] (ii)
  • Dividing equation (i) by (ii) :- [ML-1]/[MLT-2] = L-2T-2 =[T/L]2 =[TL-1]2 =1/v2
  • Therefore μ/T = 1/v2
  • v=C√T/ μ where C=dimensionless constant
  • Conclusion: v depends on properties of the medium and not on frequency of the wave.

Problem:- A steel wire 0.72 m long hasa mass of 5.0 ×10–3 kg. If the wire is under a tension of 60 N, what is the speed oftransverse waves on the wire?

Answer: -Mass per unit length of the wire,0.72 m

μ = 5.0×10−3 kg

= 6.9 ×10–3 kg m–1

Tension, T = 60 N

The speed of wave on the wire is given by,v=C√T/ μ

√ (60)/ (6.9×10-3kgm-1) = 93ms-1

Problem:- A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?


Mass of the string, M = 2.50 kg

Tension in the string, T = 200 N

Length of the string, l = 20.0 m

Mass per unit length,μ=M/l=2.50/20=0.125kgm-1

The velocity (v) of the transverse wave in the string is given by the relation: v=√T/μ


Therefore, time taken by the disturbance to reach the other end, t =l/v=20/40=0.50s

Problem:- A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.


Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.10 kg

Velocity of the transverse wave, v = 343 m/s

Mass per unit length,μ=m/l= 2.10/12 =0.175kgm-1

For tension T, velocity of the transverse wave can be obtained using the relation:

v= √T/μ

Therefore T=v2μ

= (343)2 × 0.175 = 20588.575 ≈ 2.06 × 104 N

Speed of a longitudinal wave in a stretched string

Longitudinal wave speed determined by:

  • Density – Longitudinal wave is formed due to compressions (particles very close to each other) and rarefactions (particles are far from each other).
  • At certain places it is very dense and at certain places it is very less dense.So density plays a very important role.
  • It is denoted by ρ.
  • Bulk modulus– Bulk modulus tells how does the volume of a medium changes when the pressure on it changes.
    • If we change the pressure of compressions orrarefactions then the volume of the medium changes.
    • It is denoted by B.

Dimensional Analysis to show how the speed is related to density and bulk modulus

  • ρ =mass/volume= [ML-3]
  • B = – (Change in pressure (ΔP))/Change in volume(ΔV/V))
  • ΔV/V is a dimensionless quantity as they are 2 similar quantities.
  • ΔP =F/A = ma/A =[M][LT-2]/[L2] = [ML-1T-1]
  • Dividing ρ/B= [ML-3]/[ML-1T-2] = [L-2T2]=[TL-1]2 =1/v2
  • Therefore ρ/B = 1/v2 or v2=B/ ρ
  • v= C √ (B/ ρ) where C=dimensionless constant
  • In case of fluids :- v= C √ (B/ ρ)
  • In case of solids :- v= C √ (Y/ ρ)

Problem:- Estimate the speed ofsound in air at standard temperature andpressure. The mass of 1 mole of air is29.0 ×10–3 kg.


We know that 1 mole of any gasoccupies 22.4 litres at STP. Therefore, density

of air at STP is:ρo = (mass of one mole of air)/ (volume of onemole of air at STP)

29.0 x 103 kg/22.4×103 m3

= 1.29 kg m–3

According to Newton’s formula for the speedof sound in a medium, we get for the speed of

sound in air at STP,

v = [(1.01×105Nm-2)/ (1.29kgm-3)]1/2


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