FORCED OSCILLATIONS

The phenomenon of setting a body into vibrations with the external periodic force having the

frequency different from natural frequency of body is called forced vibrations and the resulting

oscillatory system is called forced or driven oscillator.

For example, if we hold a pendulum bob in the hand, the pendulum can be given any number of swings

per second by moving hand. The pendulum oscillates with frequency of the motion of hand and not

with its own natural frequency.

At insignificant damping, when the natural frequency of the vibrating body matches (is equal to ) the

frequency of the external periodic force, body oscillates with large amplitude. These resulting

vibrations are called Resonant Force Vibrations or Resonant vibrations or Sympathetic vibrations; the

phenomenon is simply termed as Resonance. A very small periodic force may produce resonant

vibrations of extremely large amplitude if friction in the system is low. In forced vibration, the energy

drawn from the source may be either used to raise the potential or kinetic energy of the system. When

the frequency of the external periodic force is adjusted so that the amplitude of the system raises to

its maximum value, then we call such resonance as ‘Amplitude resonance’ . In this case potential

energy of the system is of maximum value. When the frequency of the external periodic force is

adjusted to increase the velocity to its maximum value, then phenomena is termed as ‘Velocity

Resonance’.

Forced oscillations When a system of mass m can execute harmonic oscillations in the presence of a mechanical resistance

R, the oscillations are called damped natural oscillations and the frequency of such oscillations is

f=1

2π√ω2 − b2 =

1

2π √

k

m−

R²

4m²

The frequency fo =1

2π√

k

m is of undamped oscillations.

If the system is subjected to a sinusoidal driving force of frequency p

2π , it executes oscillations with

the same (or very nearly the same) frequency as that of the driving force. These are called forced

oscillations or forced vibrations.

Differential equation for the motion of forced damped oscillator.

Let F = Fo sin pt or F = Focos pt or complex force Foejpt be the periodic force of frequency p/2π applied

to the damped harmonic oscillator. Then, the differential equation for the motion of the forced

damped oscillator is

md²x

dt²+ R.

dx

dt+ kx = Foe

jpt …………………………. (1)

where,

m is the mass of the system,

R is the mechanical resistance,

k is the force constant of undamped (free) oscillations and

Fo is the maximum value of the periodic force.

The solution of equation (1) is the sum of two parts:

1. A transient term containing two arbitrary constant and

2. A steady state term which depends on F and does not contain any arbitrary constant.

Transient state solution

This part of the solution is obtained by setting F = 0 in equation (1) and is given by

x= e−bt[Aeαt + Be−αt]

For damped natural oscillations,

α = j√ω² − b² = j√k

m−

R²

4m²= jβ

The frequency of damped oscillations is

f = 1

2π√

k

m−

R²

4m²

After sufficient time, such that bt >>1, the damping term e−bt makes this part of the solution

negligible. Consequently, only steady state part of the solution of frequency p

2π remains.

Steady state solution

For steady state solution, the force F0 sin pt is replaced by its equivalent complex force Fojpt

.

Then, equation (1) becomes,

md²x

dt²+ R.

dx

dt+ kx = F0e

jpt …………………………. (2)

The solution of equation (2) is of the form : x = Aejpt

where,

A, in general, is a complex quantity.

∴ dx

dt= A. jpejpt = jpAejpt = jpx

and d²x

dt²= jp.

dx

dt= jp(jpx) = j²p²x = ̶ p²x

Substituting the values of d2x

dt2,

dx

dt in equation (2), we get

(-mp² + Rjp +k)x = Foejpt

or, (jpR – p²m + k)x = Foejpt

or, [jpR – p (pm ̶ k

p)]x = Foe

jpt

Multiplying both sides by j

p [−R – j (pm ̶ k

p)]x = j Foe

jpt

or x = −j Foejpt

p[R + j(pm ̶ k

p)]

…………………………. (3)

In this equation, R + j (pm – k/p) is the complex mechanical impedance Z ⃗⃗

Thus, Z ⃗⃗ = R + j (pm −k

p) = Zejϕ …………………………. (4)

Where,

Z is the magnitude of Z ⃗⃗ .

The magnitude is given by

Z = √R² + (pm −k

p)2 …………………………. (5)

and its phase angle ϕ is given by

tan ϕ = pm−k/p

R …………………………. (6)

Now, putting the value of complex impedance Z ̅ in equation (3), we get,

x = − jFoejpt

pZejϕ

=− jFoej(pt−ϕ)

p√R²+(pm−k

p)2

=−j Fo

p√R²+(pm−k

p)2[cos (pt − ϕ) + j sin (pt − ϕ)]

= Fo

p√R²+(pm−k

p)2 sin (pt − ϕ)] −

j Fo

p√R²+(pm−k

p)2 cos (pt − ϕ)

………………………….(7)

1. Driving force F = Fo cos pt. Now, it is a real part of complex force Foejpt. Hence, the real part

of equation (3.7) is the steady state solution of damped oscillator.

∴ x = Fo

p√R²+(pm−k

p)2sin(pt − ϕ)

Putting for R/m = 2b and k/m = ω² in above equation, we get

x = Fo

p

p√4m2b²p2+(p2m-mω2)2

sin(pt − ϕ)

x = Fo

m√4b²p2+(p2- ω2)2sin(pt − ϕ) …………………………. (8)

x = Fo

m√4b²p2+(p2- ω2)2 cos [pt − (ϕ +

π

2)] …………………………. (9)

2. Driving force F=Fo sin pt. Now, it is a imaginary part of complex force Foejpt. Hence, the

imaginary part of equation (7) is the steady state solution of damped oscillator.

∴ x = − Fo

p√R²+(pm−k

p)2cos(pt − ϕ)

Putting for R/m = 2b and k/m = ω² in above equation, we get

x = − Fo

p

p√4m2b²p2+(p2m-mω2)2

cos(pt − ϕ)

x = − Fo

m√4b²p2+(p2- ω2)2cos(pt − ϕ)

x = − Fo

m√4b²p2+(p2- ω2)2 sin [pt − (ϕ +

π

2)] …………………………. (10)

From equations (9) and (10), following conclusions are made:

1. The displacement x varies with time as a sine or cosine curve with the same frequency p/2π

as that of the periodic force.

2. The amplitude of the forced S.H.M is

A = Fo

m√4b²p2+(p2- ω2)2 …………………………. (10)

Case I: At p << ω (i.e. driving frequency much less than natural frequency)

When ω ⟶0, p2 can be neglected as compared to ω2, 4b²p2 as compared to (ω2)2 and

thus A becomes, A = Fo

mω2 = Fo

k

Thus, at smaller frequencies, the amplitude of forced oscillation depends upon driving

force Fo and force constant k but independent of frequency.

Case II: At p >> ω (i.e. driving frequency very much greater than natural frequency)

For large values of p, ω2 can be neglected as compared to p2 and thus A becomes,

A = Fo

p2 (∵ b is supposed small)

Thus, as p⟶∞, p⟶0 i.e. amplitude approaches zero as driving force frequency is very

large.

Case III: At p = ω (i.e. driving force frequency is equal to natural frequency of oscillator)

This is the case of amplitude resonance. At this stage the amplitude is given by

Amax = Fo

2mbp

Thus, at resonance, maximum amplitude is inversely proportional to the damping

coefficient b.

The variation of amplitude with frequency of driving force is shown in Fig. 1

3. The displacement lags behind the periodic force by the phase angle θ, where, Ѳ = π

2+ ϕ .

Such that, tan θ = ̶ cot ϕ

= ̶ R

pm−k

p

=Rp

k−p2m =

Rp/mk

m−p²

tan θ =2bp

ω2−p2

where,

R/m = 2b and k/m = ω²

Amplitude (Displacement) resonance in forced oscillations

The amplitude of the steady-state forced oscillator is

A = Fo

m√4b²p2+(p2- ω2)2

where,

p is angular frequency of driven periodic force

ω is natural frequency of undamped oscillator

m is the mass of the oscillator

b=R/2m (R- damping constant)

Thus, the amplitude varies with the frequency f = p/2π of the applied or driven periodic force. There

exists some frequency at which amplitude of forced oscillation becomes maximum.

This phenomenon where for a particular driving frequency, the amplitude of driven oscillator is the

maximum is called amplitude resonance and particular frequency of driving force is called

resonance frequency .

Taking log of above equation, we get

log eA = logeF0 − log m −1

2loge[4b²p

2+(p2- ω2)2]

Differentiating with respect to p, we get

1

A.∂A

∂P= −

1

2[8pb²+2(p2−ω2)(2p)

p²R²+(p2m−k)²]

At amplitude resonance i.e. at p= pr, ∂A

∂p= 0

∴ 4b² + 2pr2 − 2ω2 = 0

or pr2 = ω2 − 2b²

or pr = √ω² − 2b² …………………………. (11)

or pr = √k

m−

R²

2m² (∵ ω = √

k

m and b =

R

2m)

Thus, the resonance frequency for displacement resonance is

fr =pr

2π

=1

2π√ω² − 2b² =

1

2π√

k

m−

R²

2m²

The frequency f of the damped S.H.M is

f =1

2π√ω² − b² =

1

2π√

k

m−

R²

4m²

Thus, the resonant frequency of the forced oscillator is less than natural frequency of oscillator and

also less than that of the damped oscillator. .

At the resonance frequency pr the maximum amplitude is given by

Amax =Fo

m√4b²p𝑟2+(p𝑟

2− ω2)2

Putting pr = √ω² − 2b² in above equation, we get

∴ Amax =Fo

m√4b²(ω²−2b²) +(ω²−2b²− ω2)2

=Fo

2mb√ω2−b2

Since, b = R

2m

Amax = F0

R√ω2−b2 …………………………. (12)

The variation of amplitude A with frequency of driving force p is shown in Fig. 1.

It is seen that when R = 0. Amax = ∞. When R is very small compared with k, then the resonance

frequency, fr =ω

2π= the natural frequency of undamped oscillations

and Amax =Fo

Rω

=F0m

2(R

2m)ω

=F0m

2bω

=fo

2bω where,

fo is the applied force per unit mass.

Thus, at resonance, the amplitude of displacement is affected due to damping.

Sharpness of resonance

The displacement amplitude of the forced oscillator is given by

A = Fo

m√4b²p2+(p2- ω2)2

where,

p is angular frequency of driven periodic force

ω is natural frequency of undamped oscillator

m is the mass of the oscillator

b=R/2m (R- damping constant)

Thus, at very high and very low frequencies of the driven force, the amplitude of forced oscillator

approach zero. But, at resonance frequency, pr = √ω² − 2b² , the maximum amplitude of vibration

becomes, Amax = F0

R√ω2−b2 where, b is damping coefficient. As the frequency of driven force is

increased or decreased from its resonant value pr, amplitude always falls.

The resonance is said to be sharp if the amplitude of vibration at resonance falls off quickly for a small

change in frequency on either side of resonant

frequency pr. If there is rapid fall in amplitude

from its resonant value, the sharper will be the

resonance.

The sharpness of resonance depends on the

damping.

1. When there is no damping (i.e. b = 0), the

amplitude at resonance becomes infinite as

shown by dotted curve in Fig. 1.

2. For small damping, the amplitude is large and

falls off rapidly on either side of it when

frequency of driven force differs from

resonant frequency.

3. For heavier damping (i.e. larger b or R), Fig. 1. Resonance curves

the amplitude remains more or less at

its maximum value over an appreciable range on either side of pr. The resonance in this case

is said to be flat.

Quantitatively, the sharpness of resonance is defined as the ratio of the resonant frequency to the

difference in frequencies (taken on both the sides of pr) for which the power dissipation of oscillator

becomes half the value at the resonance frequency. The point at which the power dissipation is half

of its value at resonance are called half power points.

The sharpness of resonance can be given by the expression.

S = fr

f1−f2

where,

f1 and f2 are lower and upper cut-off frequencies.

Thus, it is clear that the sharpness of resonance is dependent of damping; smaller the

damping, the sharper the resonance.

Power dissipated by forced oscillator When a damped oscillator is acted by an external driving periodic force F = Fo cos pt, then the

displacement is given by

x = Fo

p Zsin(pt − ϕ) …………………………. (13)

where,

Z = √R² + (pm −k

p)2

(Mechanical impedance)

P is the frequency of driving force

R is damping coefficient

k is the force constant

tan ϕ = pm−k/p

R

Now, the power supplied by the driving force is not stored in the mechanical oscillator but it is

used to do the work against the dissipative frictional forces.

The rate of doing work by oscillator

against the frictional forces = Frictional force ⨯ instantaneous velocity

= Rdx

dt×

dx

dt = R(

dx

dt)2

= R [p Fo

p Zcos (pt − ϕ)]

2

= R F𝑜

2

Z2 cos2 (pt − ϕ)

The average loss of power (dissipation)over a period is given by

Pav = 1

T∫ R

Fo2

Z2 cos2 (pt − ϕ)dtT

0

= R

T

Fo2

Z2 ∫1

2[1 − cos 2(pt − ϕ)]dt

T

0

= 1

2

Fo2

Z2 R

But, cos ϕ = R

Z

∴ Pav = 1

2

Fo2

Zcos ϕ …………………………. (14)

The above equation represents the average loss of power to do the work against the dissipative

frictional forces and the term cos ϕ is called as the power factor.

Power supplied by the driving force Let an external driving periodic force F = Fo cos pt acting on a damped oscillator causing a

instantaneous displacement x = Fo

p Zsin(pt − ϕ) .

∴ The velocity of the forced oscillator is given by

v = Fo

Zcos (pt − ϕ)

where,

Z = √R² + (pm −k

p)2

(Mechanical impedance)

P is the frequency of driving force

R is damping coefficient

k is the force constant

tan ϕ = pm−k/p

R

The instantaneous power supplied to the oscillator is

P =dW

dt =

Fdx

dt = F.v

i.e. p = Instantaneous force ⨯ instantaneous velocity

= Fo · Fo

Zcos (pt − ϕ)

∴ Work done in time dt = F𝑜

2

Zcos pt ∙ cos (pt − ϕ)

The work done in complete time period T = ∫ Fo

2

Zcos pt ∙ cos (pt − ϕ)

T

0dt

∴ Average power over one complete cycle is given by

Pav = 1

𝑇

Fo2

Z∫ cos pt ∙ cos (pt − ϕ)

T

0dt

On solving, the average power supplied by the driving force to the oscillator is

Pav = 1

2

Fo2

Zcos ϕ …………………………. (15)

From equation (14) and (15) it is clear that the average power supplied by the driving force

to the oscillator is equal to the average power dissipated against the frictional forces.

Resonance absorption bandwidth Let an external driving periodic force F = Fo cos pt acting on a damped oscillator.

The average power supplied by the driving force to the oscillator is given by

Pav = 1

2

Fo2

Zcos ϕ

∵ cos ϕ = r

Z ∴ Pav =

Fo2R

2 𝑍2 …………………………. (16)

where,

Z = √R² + (pm −k

p)2

P is the frequency of driving force

R is damping coefficient

k is the force constant

The power supplied depends upon the frequency p of the applied force. Pav is maximum when cos

ϕ = 1 which gives (pm −k

p) = 0 so that ϕ =0 and Z = R.

Thus, when (pm −k

p) = 0, p2 =

k

m= ω2 i.e. p = po = ω which is the case of velocity resonance.

Therefore, Pav has a maximum value at velocity resonance as the velocity and the applied force

are always in phase.

∴ Pav(max) = Fo2

2R …………………………. (17)

The variation Pav with the frequency p of driving force is shown in Fig. 2. Pav has maximum value

for p = ω and its value decreases on either side of the maximum.

The bandwidth of the resonance curve is defined as the difference in angular frequencies (taken on both

the sides of pr) for which the power dissipation of oscillator becomes half the value at the resonance

frequency. The point at which the power dissipation is half of its value at resonance are called half power

points.

Let p1 and p2 (such that p1 < po < p2 ) be the frequencies for which Pav = 1

2 Pav(max)

Therefore, from equation (16) and (17), we get

Fo2R

2 Z2 =1

2∙

Fo2

2R

∴ Z2 = 2R2

i.e. R² + (pm −k

p)2= 2R2

∴ (pm −k

p)2= R2

∴ pm −k

p= ±R

Let po be the resonant frequency, then

p2m −k

p2= +R

or p22 −

k

m=

R

mp2 ……….(i)

and p1m −k

p1= − R

or p12 −

k

m= −

R

mp1 ……….(ii) Fig. 3.2 Resonance curve

Subtracting (ii) from (i), we get

p22 − p1

2 =R

m(p2 + p1)

p2 ̶ p1 =R

m

Hence, Band width (BW), ∆p = p2 ̶ p1 =R

m

Thus, the band width is dependent of the damping coefficient.

Quality factor (Q) of forced oscillator Quality factor Q is defined as 2π times the ratio of the energy stored to the energy dissipated per

period.

Thus, Q = 2πAverage energy stored

energy lost per period =

2πEav

T×Pav …………………………. (18)

Let an external driving periodic force F = Fo cos pt acting on a damped oscillator causing a

instantaneous displacement x = Fo

p Zsin(pt − ϕ) and velocity v =

Fo

Zcos (pt − ϕ)

where,

Z = √R² + (pm −k

p)2

(Mechanical impedance)

P is the frequency of driving force

R is damping coefficient

k is the force constant

tan ϕ = pm−k/p

R

The instantaneous energy = Kinetic energy + Potential energy at any time.

E = 1

2mv2 +

1

2kx2

∴ E = 1

2m ∙

F𝑜2

Z2 cos2 (pt − ϕ) +1

2k ∙

F𝑜2

p2 Z2 sin2(pt − ϕ)

The average value of total energy over a period is given by

Eav =1

T∙1

2∙Fo

2

Z2 [m∫ cos2 (pt − ϕ)dt +k

p2 ∫ sin2(pt − ϕ)dtT

0

T

0]

Since, 1

𝑇 ∫ cos2 (pt − ϕ)dt =

1

𝑇 ∫ sin2(pt − ϕ)dt

T

0

T

0=

1

2

∴ Eav =1

2∙Fo

2

Z2 (m

2+

k

2p2)

Since, k

m= ω2 = p𝑜

2

∴ Eav =1

4∙ m ∙

Fo2

Z2 (1 +p𝑜

2

p2)

Now, the average energy lost, Pav = Fo2R

2 𝑍2

Substituting the values of Eav and Pav in equation (18), we get

Q = 2π

1

4∙m∙

Fo2

Z2(1+p𝑜2

p2)

T×Fo2R

2 𝑍2

= π

T

m

R(1 +

po2

p2)

At resonance, p = po =2π

T

∴ Q = mpo

R

But, Band width (BW), ∆p = p2 ̶ p1 =R

m

∴ Q = mpo

R =

po

p2 ̶ p1 =

fo

f2 ̶ f1 …………………………. (19)

Thus, the Q value of a oscillator is also defined as the ratio of frequency at velocity resonance to the

full bandwidth at half maximum power. It is also called as figure of merit.

Q factor is a measure of the sharpness of the resonance curve. Larger the value of Q, sharper the

resonance absorption curve. Q value also measures the amplification factor. At displacement

resonance the displacement at low frequency is amplified by Q times.

Solved examples 1. A damped oscillator consists of a mass 200 gm attached to a spring of constant 100Nm−1 and

damping constant 5 Nm−1s. It is driven by a force F = 6 cos wt Newton, where ω = 30s−1. If

displacement in steady state is x= A sin (ωt – ϕ) metre, find A and ϕ. Also calculate the power

supplied to the oscillator.

Solution: Given, Amplitude of the driving force Fo = 6N

Frequency of the driving force p = 30s−1

Mass of damped oscillator m = 200 gm = 0.2 kg

Spring constant k = 100 Nm−1

Damping constant R = 5 Nm−1s

When an external force F = Focos ωt acts on a damped oscillator, the amplitude in steady

state is given by

A = F0

p Zm

where,

Zm being the ‘impendance’ of the mechanical system

and Zm = √R² + (pm −k

m)²

and tan ϕ = pm−

k

m

R

∴ tan ϕ = 30×0.2−

100

30

5= 0.534

∴ ϕ = 28°6’ and cos ϕ = 0.8821

Zm = √(5 × 5) + (30 × 0.2 −100

30)²

= 5.67 Nm−1s

2. If the resonant (angular) frequency of acoustic system is 280 Hz and half power frequencies are

200 Hz and 360 Hz respectively, calculate the quality factor.

Solution: Given- fo= 280 Hz, f1 = 200Hz f2= 360 Hz

∴ Quality factor, Q = f0

f2−f1

=280

360−200

= 280

160

= 1.75

3. The equation of motion is 2 x 10−4 d²x

dt²+ 4 × 10−2.

dx

dt+ 5x = 0.124 sin 100t where, all quantities

are in S.I units. Find (i) Natural frequency of undamped oscillation (ii) Mechanical impedance.

Solution: Given- differential equation to forced oscillator is

2 x 10−4 d²x

dt²+ 4 × 10−2.

dx

dt+ 5x = 0.124 sin 100t

Comparing with general equation m d²x

dt²+ R

dx

dt+ kx = F0sin pt, we get,

m = 2 x 10−4 kg, R = 4 x 10−2Nsm−1, k = 5Nm−1, Fo = 0.124 N ,

and p = 100 rad s−1

(i) Natural frequency of undamped oscillation is

Fo =1

2π√

k

m

= 1

2×3.14√

5

2×10−4 =102

6.28√

5

2= 25.16Hz

(ii) Mechanical impedance (Z)

Amplitude, A = Fo

m√4b²p2+(p2−ω2)²=

F0

pZ

∴ Mechanical impedance, Z = m

p√4b²p2 + (p2 − ω2)²

Here , ω2 − p2 =5

2×10−4 − 1002

= 1.5 × 104

and 4b²p2 = (2b)²p2= (R

m)2p2 = (

4×10−2

2×10−4)2

× (102)2 = 4 × 108

∴ Z =2 × 10−4

102√4 × 108 + 2.25 × 108 = 2 × 10−6√108(4 + 2.25)

= 2 x 10−6 × 104 × √6.25

= 2 x 2.5 x10−2

= 0.05 Nsm−1

4. A harmonic oscillator consisting of 50 gm mass attached to a mass less spring has a quality factor

200. If it oscillates with an amplitude of 2 cm in resonance with a periodic force of frequency 20

Hz. If it oscillates with amplitude of 2 cm in resonance with a periodic force of frequency 20Hz.

Calculate (i) the average energy stores in it and (ii) the rate of dissipation of energy.

Solution: The average energy stored in the oscillator is equal to its maximum potential energy

= 1

2kx² =

1

2mω2x0

2

Given that m = 50 gm, ω= 2πn = 2π x 20 = 40π, and x0= 2 cm

(i) The average energy stored or absorbed in the oscillator

= 1

2× 50 × (40π)2 × (2)²

= 1.58 x 106ergs

(ii) The quality factor

Q = 2π Average energy stored

Energy dissipated per cycle

∴ Energy dissipated per cycle = 2π x Average energy stored

Q

= 2π x 1.58×106

200= π × 1.58 × 104 Ergs

Since, there are 20 cycles per second, we have energy dissipated per second or rate of

dissipated of energy = 20 x π x 1.58 x 104

= 9.926 x 105

= 106 erg/sec

Exercise 1. A damped harmonic oscillator is subjected to a sinusoidal driving force whose frequency is

altered but amplitude kept constant. It is found that the amplitude of the oscillator increases

from 0.02 mm at a very low driving frequency to 8.0 mm at a frequency of 100 cps. Obtain the

values of (i) the quality factor, (ii) the relaxation time, (iii) the damping factor and (iv) the half-

width of the resonance curve.

1. A damped harmonic oscillator of quality factor 20 is subjected to a sinusoidal driving force of

frequency twice the natural frequency of the oscillator. If the damping the small, what fraction

will the amplitude of the oscillator be of its maximum value and by what angle will it differ in

phase from the driving forces.