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 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.

Kinetic energy, potential energy and the total energy is a function of displacement in the above graph.

The kinetic energy (K) of a particle executing SHM can be defined as

K= ½ mv2

= ½ mω2A2sin2 (ωt + φ)

K=½ k A2 sin2 (ωt + φ)

  • The above expression is a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean position.

The potential energy (U) of a particleexecuting simple harmonic motion is,

U(x) = ½ kx2

U½ k A2 cos2 (ωt + φ)

  • The potential energy of a particle executing simple harmonic motion is alsoperiodic, with period T/2, being zero at the mean position and maximum at the extremedisplacements.

Total energy of the system always remains the same

E = U + K

= ½ k A2 sin2 (ωt + φ) + ½ k A2 cos2 (ωt + φ)

E=½ k A2(sin2 (ωt + φ) + cos2 (ωt + φ))

The above expression can be written as

½ k A2

Total energy is always constant.

Problem: –A block whose mass is 1 kgis fastened to a spring. The spring has a

spring constant of 50 N m–1. The block ispulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionlesssurface from rest at t = 0. Calculate thekinetic, potential and total energies of theblock when it is 5 cm away from mean position?

Answer: -The block executes SHM, its angularfrequency, according to equation, ω= √k/m

= √ (50 N m–1)/ 1kg

= 7.07 rad s–1

Its displacement at any time t is then given by,

x (t) = 0.1 cos (7.07t)

Therefore, when the particle is 5 cm away fromthe mean position, we have

0.05 = 0.1 cos (7.07t)

Or cos (7.07t) = 0.5 and hence

sin (7.07t) = √3/2= 0.866

Then, the velocity of the block at x = 5 cm is

 = 0.1 7.07 0.866 m s–1

 = 0.61 m s–1

Hence the K.E. of the block,

=1/2 m v2

 = ­[1kg (0.6123 m s–1 )2 ]

 = 0.19 J

The P.E. of the block,

=1/2 kx2

 = ­ (50 N m–1 0.05 m 0.05 m)

 = 0.0625 J

The total energy of the block at x = 5 cm,

 = K.E. + P.E.

 = 0.25 J

we also know thatmaximum displacement,K.E. is zero and hence the total energy of thesystem is equal to the P.E. Therefore, the totalenergy of the system,

 = ­ (50 N m–1 0.1 m 0.1 m)

 = 0.25 J , which is same as the sum of the two energies ata displacement of 5 cm.This shows the result with the accordance of conservation of energy.

Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
  • T = 2√m/k where T is the period.

Problem: – A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Answer:

Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg

where,g = acceleration due to gravity = 9.8 m/s2

F = 50 × 9.8 = 490 N

∴spring constant, k = F/l = 490/0.2 = 2450 N m-1.

Mass m, is suspended from the balance  

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N

Hence, the weight of the body is about 219 N.

Problem: – A 5 kg collar is attached to a spring of spring constant 500 N m–1. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate

(a) the period of oscillation,

(b) the maximum speed and

(c) maximum acceleration of the collar.

Answer:-

The period of oscillation as given by

T =2√m/k = 2 π√ (5.0 kg/500 Nm-1)

 = (2π/10)s

= 0.63 s

(b) The velocity of the collar executing SHM is

given by,

v (t) = –Aω sin (ωt + φ)

The maximum speed is given by,

vm = Aω

 = 0.1√ (k/m)

= 0.1√ (500Nm-1/5kg)

= 1 ms–1

and it occurs at x = 0

(c) The acceleration of the collar at thedisplacement x (t) from the equilibrium isgiven by,

a (t) = –ω2x (t)

 = – k/m x (t)

Therefore the maximum acceleration is,

 amax = ω2 A

 = (500 N m–1/5kg) x0.1m

= 10 m s–2and it occurs at the extreme positions.

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