Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution

www.free-education.in is a platform where you can get pdf notes from 6th to 12th class notes, General Knowledge post, Engineering post, Career Guidelines , English Speaking Trick , How to crack interview and lots more. ( Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution )

Class 11 Physics Chapter 14 Oscillations NCERT Solution

free education
Section NameTopic Name
14Oscillations
14.1Introduction
14.2Periodic and oscilatory motions
14.3Simple harmonic motion
14.4Simple harmonic motion and uniform circular motion
14.5Velocity and acceleration in simple harmonic motion
14.6Force law for simple harmonic motion
14.7Energy in simple harmonic motion
14.8Some systems executing Simple Harmonic Motion
14.9Damped simple harmonic motion
14.10Forced oscillations and resonance
Oscillations Notes

Introduction

In this chapter we will learn about oscillatory motion or oscillations. Any motion which repeats itself at regular intervals of time is known as periodic motion. If a body moves back and forth repeatedly about its mean position then it is said to be in oscillatory motion.

For example: The to and fro movement of pendulum, jumping on a trampoline, a child swinging on a swing.

Oscillations can be defined as Periodic to and fro motion which repeat itself at regular intervals of time.

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

Periodic and Oscillatory motions

Oscillations are defined as to and fro motion which repeat itself after regular intervals of time.In oscillations, the frequency of vibrations iscomparatively less. 

For example: The to and fro motion of a pendulum clock

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

Frequency

It is defined as number of cycles per second.

  • It is denoted by ν.
  • I.unit is sec-1
  • Special Unit is Hertz(Hz)
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.

Problem: – On an average human heart is found to be beat 75 times in a minute. Calculate frequency and time period?

Answer: –

The beat frequency of heart = 75/ (1 min)

 = 75/ (60 s)

 = 1.25 s–1

 = 1.25 Hz

 The time period T = 1/ (1.25 s–1) = 0.8 s 

Oscillations Notes

In the above image we can see that there is a block whose one end is attached to a spring and another is attached to a rigid wall.x is the displacement from the wall.

In the above figure a blockis attached to a spring, the other end of which is fixed to a rigid wall. The block moves on a frictionless surface. The motion of the block can be described in terms of its distance or displacement x from the wall.

f (t) = A cos ωt

As cosine function repeats after 2π so it can be written as

cos (θ) = cos (ωt + 2π)            Equation (1)

cos (ωt) = cos (ωt + 2π) (it keep on repeating after 2π)

Let Time Period = T

f (T) = f(t+T) where displacement keeps on repeating after (t+T)

Acos (ωt) = cosω(t+T) = Acos (ωt+ wT)

Acosωt = A cos (ωt+ωT)          Equation (2)

From Equation (1) and Equation (2)

ωT= 2π

Or T=2π /ω

Displacement as a combination of sine and cosine functions

f (t) = A cos ωt

f (t) = A sin ωt

f (t) = A sin ωt + A cos ωt

LetA = D cosΦ              Equation (3)

B=DsinΦEquation (4)

f (t) =DcosΦ sinωT + DsinΦ cos ωt

D (cosΦ sinωT + sinΦ cos ωt)

(Using sinAcosB + sinBcosA = sin (A+B))

Therefore we can write

f (T)= D sin (ωT+Φ)

From the above expression we can say displacement can be written as sine and cosine functions.

D in terms of A and B:-

AB= D2sin2 Φ + D2cos2 Φ

AB= D2

Or D= AB

Φ In terms of A and B

Dividing Equation (4) by (3)

B/A= DsinΦ/Dcos Φ

tan Φ = B/A

Or Φ= tan-1 B/A

Problem:-Which of the followingfunctions of time represent (a) periodic and (b) non-periodic motion? Give the period foreach case of periodic motion [ω is anypositive constant].

(i) sin ωt + cos ωt

(ii) sin ωt + cos 2 ωt + sin 4 ωt

(iii) e–ωt

(iv) log (ωt)

 Answer:-

  • sin ωt + cos ωt is a periodic function, it can also be written as

 2 sin (ωt + π/4).

Now 2 sin (ωt + π/4)= 2 sin (ωt + π/4+2π)

= 2 sin [ω (t + 2π/ω) + π/4]

The periodic time of the function is 2π/ω.

(ii) This is an example of a periodic motion. Itcan be noted that each term represents aperiodic function with a different angularfrequency. Since period is the least intervalof time after which a function repeats itsvalue, sin ωt has a period T0= 2π/ω; cos 2 ωt

has a period π/ω =T0/2; and sin 4 ωt has aperiod 2π/4ω = T0/4. The period of the firstterm is a multiple of the periods of the last

two terms. Therefore, the smallestintervalof time after which the sum of the threeterms repeats is T0, and thus the sum is aperiodic function with aperiod 2π/ω.

(iii) The function e–ωt is not periodic, itdecreases monotonically with increasingtime and tends to zero as t →∞ and thus,never repeats its value.

(iv) The function log (ωt) increases monotonicallywith time t. It, therefore, neverrepeats its value and is a non-periodicfunction. It may be noted that as t →∞,log (ωt) diverges to ∞. It, therefore, cannot represent any kind of displacement.

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

Phase

It is that quantity that determines the state of motion of the particle.

  • Its value is (ωt + Φ)
  • It is dependent on time.

Value of phase at time t=0, is termed as Phase Constant. When the motion of the particle starts it goes to one of the extreme position at that time phase is considered as 0.

Let x (t) = A cos (ωt) where we are taking (Φ = 0)

  1. Mean Position (t= 0)
  2. x (0) = A cos (0) = A (cos0=1)
  3. t=T/4, t= T/2, t=3T/4, t=T and t=5T/4
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Oscillations Notes

Answer:

In SHM, acceleration a is related to displacement by the relation of the form

 a = -kx, which is for relation (c).

Problem: – The motion of a particle executing simple harmonic motion is described by the displacement function,

x (t) = A cos (ωt + φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Answer:-

Initially, at t = 0;

Displacement, x = 1 cm

Initial velocity, v = ω cm/ sec.

Angular frequency, ω = π rad/s–1

It is given that,

x (t) = A cos (ωt + Φ)

1 = A cos (ω × 0 + Φ) = A cos Φ

A cosΦ = 1                          … (i)

Velocity, v= dx/dt

ω = -A ω sin (ωt + Φ)

1 = -A sin (ω × 0 + Φ) = -A sin Φ

A sin Φ = -1                         … (ii)

Squaring and adding equations (i) and (ii), we get:

A2 (sin2 Φ + cos2 Φ) = 1 + 1

A2 = 2

∴ A = √2 cm

Dividing equation (ii) by equation (i), we get:

tan Φ = -1

∴Φ = 3π/4, 7π/4…

SHM is given as:

x = B sin (ωt + α)

Putting the given values in this equation, we get:

1 = B sin [ω × 0 + α] = 1 + 1

B sin α = 1                           … (iii)

Velocity, v = ω B cos (ωt + α)

Substituting the given values, we get:

π = π B sin α

B sin α = 1                            … (iv)

Squaring and adding equations (iii) and (iv), we get:

B2 [sin2 α + cos2 α] = 1 + 1

B2 = 2

∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get:

B sin α / B cos α = 1/1

tan α = 1 = tan π/4

∴α = π/4, 5π/4…

Angular Frequency (ω)

Angular frequency refers to the angular displacement per unit time. It can also be defined as the rate of change of the phase of a sinusoidal waveform (e.g., in oscillations and waves). Angular frequency is larger than frequency ν (in cycles per second, also called Hz), by a factor of 2π.

Consider the oscillatory motion which is varying with time t and displacement x of the particle from the origin:

x (t) = cos (ωt + Φ)

Let Φ = 0

x (t) = cos (ωt)

After t=T i.e. x (t) = x (t+T)

A cos ωt = A cos ω (t + T)

Now the cosine function is periodic with period 2π, i.e., it first repeats itself after 2π. Therefore,

ω (t + T) = ωt + 2π

i.e. ω = 2π/ T

Where ω = angular frequency of SHM.

  • I. unit is radians per second.
  • It is 2π times the frequency of oscillation.
  • Two simple harmonic motions may have the same A and φ, but different ω.

Problem: – Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(a) sin ωt – cos ωt

(b) sin3ωt

(c) 3 cos (π/4 – 2ωt)

(d) cos ωt + cos 3ωt + cos 5ωt

(e) exp (–ω2t2)

Answer: –

(a) SHM

The given function is:

sinωt – cos ωt

This function represents SHM as it can be written in the form: a sin (ωt + Φ)

Its period is: 2π/ω

(b) Periodic but not SHM

The given function is:

sin 3ωt = 1/4 [3 sin ωt – sin3ωt]

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

Its period is: 2π/ω

(c) SHM

The given function is:

This function represents simple harmonic motion because it can be written in the form: a cos (ωt + Φ) its period is: 2π/2ω = π/ω

(d) Periodic, but not SHM

The given function is cosωt + cos3ωt + cos5ωt. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic. 

(e) Non-periodic motion

The given function exp (-ω2t2) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.

 (f) The given function 1 + ωt + ω2t2 is non-periodic.

ProblemWhich of the followingfunctions of time represent (a) simple harmonic motion and (b) periodic but notsimple harmonic? Give the period for each case?

(1) sin ωt – cos ωt

(2) sin2 ωt

Answer:

(a) sin ωt – cos ωt

= sin ωt – sin (π/2 – ωt)

= 2 cos (π/4) sin (ωt – π/4)

 = √2 sin (ωt – π/4)

This function represents a simple harmonicmotion having a period T = 2π/ω and aphase angle (–π/4) or (7π/4).

(b) sin2 ωt

= ½ – ½ cos 2 ωt

The function is periodic having a periodT = π/ω. It also represents a harmonicmotion with the point of equilibriumoccurring at ½ instead of zero.

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

As the particle is moving in the same way the projections are also moving.

  • When the particle is moving in the upper part of circle then the projections start moving towards left.
  • When the particle is moving in the lower part of the circle then the projections are moving towards right.
  • We can conclude that the particle is swinging from left to right and again from right to left.
  • This to and fro motion is SHM.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Oscillations Notes

Answer: – (a) Time period, t = 2 s

Amplitude, A = 3 cm

At time, t = 0, the radius vector OP makes an angle π/2 with the positive x-axis, phase angle Φ = +π/2

Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given by the displacement equation:

                 A= cos [(2 πt/T) + Φ]

                 =3cos (2 πt/2 + π/2) = -3sin (2πt/2)

                 =-3sinπt cm.

(b) Time Period, t = 4 s

Amplitude, a = 2 m

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction, Hence, phase angle Φ = +π

Therefore, the equation of simple harmonic motion for the x-projection of OP, at the time t, is given as:

                 =a cos [(2 πt/T) + Φ]

                 =2 cos [(2 πt/T) + π]

                 x=-2 cos (π/2 t) m

Velocity in Simple Harmonic Motion

  • Uniform Circular motion can be defined as motion of an object in a circle at a constant speed.
  • Consider a particle moving in circular path
  • The velocity at any point P at any time t will be tangential to the point P.
  • Consider θ = ωt+ φ where
  • θ = angular position
  • ω = angular velocity of the particle
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.

ap = -ω2Awhere

  • A = radius of the circle
  • (-ive sign shows it is pointing towards the centre of the circle.)
  • Consider the acceleration of the projection of the particle P’ on the x-axis.
  • Accelerationwill be given as
  • a(t) = -acosθ
  • a(t) = – ap cos(ωt + φ)
  • 2A cos (ωt + φ)
  • a(t) = – ω2x(t) (Using x(t)= A cos (ωt + φ))

 To verify expression for acceleration when calculated directly from SHM –

  • Displacement in SHM is = A cos (ωt + φ)
  • Velocity v(t) in SHM is = -Aw sin (ωt + φ)

Therefore,

  • a(t) = dv/dt
  • = – ω2A cos (ωt + φ)
  • a(t) = – ω2x(t) (Using x(t)= A cos (ωt + φ))

Equation of acceleration of the particle which executesSHM:-

a(t) = – ω2x(t)  

We can conclude that:-

  1. a is proportional to displacement
  2. acceleration is always directed towards the centre(in circular motion centre is mean position of the SHM)

From above we can say that

  • SHM is the projection of the uniform circular motion such that centre of uniform circular motion becomes the mean position of the SHM and the radius of the circular motion is the amplitude of the SHM.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.

Problem: -A body oscillates with SHM according to the equation (in SI units),   x = 5 cos [2π t + π/4]

At t = 1.5 s, calculate the (a) displacement,(b) speed and (c) acceleration of the body?

Answer:

The angular frequency ω of the body

= 2π s–1

and its time period T = 1 s.

At t = 1.5 s

(a) Displacement = (5.0 m) cos [(2π s–1)1.5 s + π/4]

= (5.0 m) cos [(3π + π/4)]

= –5.0 x 0.707 m

= –3.535 m

(b) The speed of the body

= – (5.0 m)(2π s–1) sin [(2π s–1) 1.5 s+ π/4]

= – (5.0 m) (2π s–1) sin [(3π + π/4)]

 = 10π (0.707) m s–1

 = 22 m s–1

(c) The acceleration of thebody

= – (2π s–1)2displacement

 = – (2π s-1)2 (–3.535 m)

 = 140 m s–2

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

F1 = -kx (force exerted by the spring onthe left side, trying to

                pull the mass towards the mean position)

F2 = -kx   (force exerted by the spring onthe right side, trying to pull the mass towards the mean position)

The net force, F, acting on the mass is thengiven by,

F = –2kx

Therefore, the force acting on the mass is proportional to the displacement and is directedtowards the mean position; therefore, the motion executed by the mass is simple harmonic. The time period of oscillations is given as:-

T=2π√m/2k

Problem:-The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s? (g on the surface of earth is 9.8 ms–2)

Answer: -Acceleration due to gravity on the surface of moon,g’ = 1.7 m s–2

Acceleration due to gravity on the surface of earth, g = 9.8 m s–2

Time period of a simple pendulum on earth, T = 3.5 s

T=2π√l/g

where l =length of the pendulum

l= T2/ (2π) 2x g

= (3.5)2/ (4x (3.14)2) x 9.8 m

The length of the pendulum remains constant,

On moon’s surface, time period, T’= 2π√l/g’

    =2π ((3.5)2/4x(3.14)2 x 9.8)/1.7

    =8.4s

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

Problem:-

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle:

T = 2π√ (m/k). A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2π√ (l/g)

Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabinthat is freely falling under gravity?

Answer:

(a) For a simple pendulum, force constant or spring factor k is proportional to mass m; therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.

(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:

F = –mg sinθ

where,

F = Restoring force

m = Mass of the bob

g = Acceleration due to gravity

θ = Angle of displacement

For small θ, sinθ≈θ

For large θ, sinθ is greater than θ.

This decreases the effective value of g.

Hence, the time period increases as:

T = 2π√ (l/g)

(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.

(d) Gravity disappears for a man under free fall, so frequency is zero

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

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.

Simple Pendulum

A simple pendulum is defined as an object that has a small mass (pendulum bob), which is suspended from a wire or string having negligible mass.     

  • Whenthe pendulum bob is displaced it oscillates on a plane about the vertical line through the support.
  • Simple pendulum can be set into oscillatory motion by pulling it to one side of equilibrium position and then releasing it.
Class 11 Physics Chapter 14 Oscillations Notes and NCERT Solution. www.free-education.in provide study material to excel in exam.

Problem:-A cylindrical piece of cork of density of base area A and height h floats in a liquid ofdensity ρl. The cork is depressed slightly and then released. Show that the corkoscillates up and down simple harmonically with a period

 T = 2 π√hρ/ρlg

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid)

Answer:-

Base area of the cork = A

Height of the cork = h

Density of the liquid = ρ1

Density of the cork = ρ

In equilibrium:

Weight of the cork = Weight of the liquid displaced by the floating cork

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

Up-thrust = Restoring force, F = Weight of the extra water displaced

F = ­–(Volume × Density × g)

Volume = Area × Distance through which the cork is depressed

Volume = Ax

∴ F = – A x ρ1 g …..(i)

According to the force law:

F = kx

k = F/x

where, k is constant

k = F/x = -Aρ1 g…. (ii)

The time period of the oscillations of the cork:

T = 2π√m/k      …. (iii)

where,

m = Mass of the cork

= Volume of the cork × Density

= Base area of the cork × Height of the cork × Density of the cork

= Ahρ

Hence, the expression for the time period becomes:

         T=2π√Ahρ/Ahρ1g      

           T = 2π√hρ/ρ1g

DAMPED SIMPLE HARMONIC MOTION

Damped SHM can be stated as:-

  1. Motion in which amplitude of the oscillating body reduces and eventually comes to its mean position.
  2. Dissipating forces cause damping.
  3. Consider a pendulum which is oscillating
  4. After some time we can observe that its displacement starts decreasing and finally it comes to rest.
  5. This implies that there is some resistive force which opposes the motion of the pendulum. This type of SHM is known as Damped SHM.

Damping Force:-

  • It opposes the motion of thebody.
  • Magnitude of damping force is proportional to the velocity of the body.
  • It actsin the opposite direction of the velocity.
  • Denoted by Fdwhere d is the damping force.
    • Fd= -b v where b is a damping constant and it depends on characteristics of the medium (viscosity, for example) and the size and shape of the block.
  • (-ive) directed opposite to velocity

Equation for Damped oscillations: Consider a pendulum which is oscillating.

It will experience two forces

  1. Restoring force Fs = -k x
  2. Damping Force Fd = -b v

The total force Ftotal =  F+ Fd       = -k x – b v

Let a (t) = acceleration of the block

Ftotal= m a (t)

-k x – b v = md2x/dt2

md2x/dt2 + kx + bv =0

or md2x/dt2 + b dx/dt+ kx=0   (v=dx/dt) (differential equation)

d2x/dt2+ (b/m) dx/dt+ (k/m) x=0  

After solving this equation

x(t) = A e–b t/2m cos (ω′t + φ ) (Equation of damped oscillations)

Damping is caused by the term e–b t/2m

  ω’ =angular frequency

Mathematically can be given as:-

   ω ′= −√ (k/m –b2/4m2)

Consider if b=0 (where b= damping force) then

x (t) = cos (ω′t + φ)( Equation of Simple Harmonic motion)

Graphically if we plot Damped Oscillations

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

In the above figure there are set of 5 pendulums of different lengths suspended from a common rope.

  • The figure has 4 pendulums and the strings to which pendulum bobs 1 and 4 are attached are of the same length and the others are of different lengths.
  • Once displaced, the energy from this pendulum gets transferred to other pendulums through the connecting rope and they start oscillating. The driving force is provided through the connecting rope and the frequency of this force is the same as that of pendulum 1.
  • Once pendulum 1 is displaced, pendulums 2, 3 and 5 initially start oscillating with their natural frequenciesand different amplitudes, but this motion is gradually damped and not sustained.
  • Their oscillation frequencies slowly change and later start oscillating with thefrequency of pendulum 1, i.e. the frequency ofdriving force but with different amplitudes.
  • They oscillate with small amplitudes. The oscillation frequency of pendulum 4 is different than pendulums 2, 3 and 5.
  • Pendulum 4 oscillates with the same frequency as that of pendulum 1 and its amplitude gradually picks up and becomes very large.
  • This happens due to the condition for resonance getting satisfied, i.e. the natural frequency of the system coincides with that of the driving force

Related link you must like:-

HVAC Course

CBSE Study Material

Study material for Competition Exam

Interview Question

Career Option

Spoken English

Leave a Reply

Your email address will not be published. Required fields are marked *