Course Content
Class 11 Physics Chapter 1 Physical World
Section Name Topic Name 1 Physical World 1.1 What is physics? 1.2 Scope and excitement of physics 1.3 Physics, technology and society 1.4 Fundamental forces in nature 1.5 Nature of physical laws
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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
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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
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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
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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
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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
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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
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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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Class 11th Physics Online Class For 100% Result
About Lesson

Modes of Oscillations:-

  • ν= (vn)/2L where v=speed of the travelling wave, L=length of the string,  n=any natural number.
  • First Harmonic:-
    • For n=1, mode of oscillation is known as Fundamental mode.
    • Therefore ν1=v/(2L).This is the lowest possible value of frequency.
    • Therefore ν1is the lowest possible mode of the frequency.
    • 2 nodes at the ends and 1 antinode.
  • Second Harmonic:-
    • For n=2, ν2=(2v)/ (2L) =v/L
    • This is second harmonic mode of oscillation.
    • 3 nodes at the ends and 2 antinodes.
  • Third Harmonic:-
    • For n=3,ν= (3v)/ (2L).
    • This is third harmonic mode of oscillation.
    • 4 nodes and 3 antinodes.

Problem:- Find the frequency of note emitted (fundamental note) by a string 1m long and stretched by a load of 20 kg, if this string weighs 4.9 g. Given, g = 980 cm s–2?

Answer:-

L = 100 cm T = 20 kg = 20 × 1000 × 980 dyne

m= 4.9/100 = 0.049 g cm-1

Now the frequency of fundamental note produced,

ν= (1/2L) √ (T/m)

ν = 1/ (2×100) √(20x1000x980)/(0.049)

=100Hz

Problem:- A pipe 20 cm long is closed at one end, which harmonic mode of the pipeis resonantly excited by a 430 Hzsource? Will this same source can bein resonance with the pipe, if both ends are open? Speed of sound = 340ms–1?

Answer:-

The frequency of nth mode of vibration of a pipe closed at one endis given by

νn=(2n 1)ν/4L

ν=340ms-1; L=20cm=0.2m;νn=430Hz.

Therefore 430= ((2n-1) x 340)/ (4×0.2)

=>n=1

Therefore, first mode of vibration of the pipe is excited, for open pipe sincen must be an integer, the same source cannot be in resonance with thepipe with both ends open.

Nodes and Antinodes: system closed at one end

  • For a system which is closed at one end,only one node is formed at the closed end.
  • Consider there is one fixed end(x=0) and one open end(x=L), and a string is attached between these two ends.
  • At x=L, antinodes will be formed.This means amplitude will be maximum at this end.
  • Condition for formation of antinodes is x= (n+ (1/2)) (λ/2).=>L=(n+ (1/2)) (λ/2)
  • =>λ=(2L)/(n+(1/2)).This expression shows that the values of wavelength is restricted. n=0, 1, 2, 3…
  • Corresponding frequencies will be ν= v/(2L)(n+1/2) (By using ν=v/λ);n=0,1,2,3, …

Modes of oscillations:-

  • Fundamental frequency:- Also known as First Harmonic:-It corresponds to lowest possible value for n. That is n=0.
  • The expression for fundamental frequency is ν0 = v/(2L)x1/2 = v/(4L)
  • Odd Harmonics
    • n=1 ; ν =v/(2L)(1+(1/2)) = (3v)/(4L) =3ν0
    • n=2; ν =v/(2L)(2+(1/2))= (5v)/(4L)=5ν0
    • n=3;ν =v/(2L)(3+(1/2))=(7v)/(4L)=7ν0
  • For a system which is closed at one end and open at another end will get one fundamental frequency and all other odd harmonics.

Therefore nx550=1100 => n=2. It is the second harmonic.

  1. Fundamental Frequency: – ν1= (V)/ (4L)

Odd harmonics :- (3V)/ (4L), (5V)/ (4L), (7V)/ (4L) ….

=>ν1Y =275Hz. Where νY = Fundamental Frequency

=>nν1 = 1100 => n=1100/275=4

This should correspond to 4th harmonic but 4th harmonic cannot be present as only odd harmonics are present.

Therefore the same source cannot be resonated if the pipe is closed at one end.

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