Superposition of Two or More Simple Harmonic Oscillators: Notes with Example | Oscillations, Waves & Optics - Physics PDF Download

Superposition of Two Collinear Harmonic Oscillations

Addition of Two S.H.M having Equal Frequencies

Consider two SHMs of equal frequencies but of different amplitudes and phase constants

acting on a particle in the x -direction.

The displacement x₁ and x₂ of the two SHM of frequency ω is

x₁ = A₁ sin (ωt +φ₁) and x₂ = A₂ sin (ωt +φ₂)

where A₁ and A₂ are the amplitude and φ₁ and φ₂ the initial phases of the two motions.

The resultant displacement x at any instant t is

x = x₁ + x₂ = A₁ sin (ωt +φ₁) = A₂ sin (ωt +φ₂)

⇒ x = A₂ sin ωt +φ₂ = A₁(sinωt cosφ₁ + cosωt sinφ₁) + A₂ (sinωt cosφ₂ + cosωtsinφ₂)

⇒ x = sinωt (A₁ cosφ₁ + A₂ cosφ₂) + cosωt (A₁sinφ₁ + A₂sinφ₂)

Now Let

A₁ cosφ₁ + A₂ cosφ₂ = Acosδ and A₁ sinφ₁ + A₂sinφ₂ = Asinδ

where A and δ are constant to be determined.

⇒ x = Asin (ωt +δ )

This shows that the resultant motion is SHM with angular frequency ω, the same as that

of the individual SHMs.

The resultant motion has amplitude A and a phase constant δ

where A =  and

Conclusion: superposition of collinear SHM is also a SHM of the same frequency but

different amplitude and phase constant

Case-I: (Maximum Amplitudes)

When the phase difference between the two individual motions is zero or any integral

multiple of 2π i.e. φ₁ −φ₂ = 2nπ n = (0,1, 2,3...)

Then A = A₁ + A₂ (Resultant amplitude is sum of the amplitudes of individual motions)

Case-I: (Minimum Amplitudes)

Whenφ1 −φ2 = (2n +1)π , n = (0,1, 2,3...) this gives A = A₁ − A₂

If A₁ = A₂ then A = 0 , i.e particle at rest.

Addition of Two S.H.M having Different Frequencies

Consider two SHM of equal amplitude but different frequencies

x₁= A₁ sinω₁ t and x₂ = A₂ sinω₂ t

The resultant displacement is

x =

This represent a periodic motion) of amplitude A =

⇒

Thus resultant amplitude of motion varies periodically between 1 ±2A and zero.

The amplitude A is maximum when

Hence the time interval between two consecutive maxima is T_b = .

The frequency ν_b of maxima is = ν₁ −v₂

The amplitude A is minimum when

or  (n = 0,1,2,3....)

or

Thus the time interval between two consecutive minima is T_b =

Hence the frequency of minima is also (ν₁ −v₂) .

One maximum of amplitude followed by a minimum is called a beat. The time period T_b

between the successive beats is called the beat period  T_b =  and beat frequency

Superposition of Two Perpendicular Harmonic Oscillations

Addition of Two SHM having Equal Frequencies

Let us consider two perpendicular SHM one along x -axis and other along y -axis with

amplitude A₁ & A₂

x = A₁ sin ωt +δ and y = A₂ sinωt where δ is phase constant.

⇒

∵ sin ωt = y/A₂ and cosωt =

⇒

This is the general equation of ellipse whose axes are inclined to the co-ordinate axes.

Let us consider few cases

(i) δ = 0 ⇒

or

This is the equation of straight line having a positive slope A₂/A₁ and passing through the origin.

Motion description: x = A₁ sinωt and y = A₂ sinωt

At time t = 0 , particle is at O, at t = , particle is at M .

At t = , =π , particle is at O, at t = , particle is at M′ .

and at t = T (ωt = 2π ) , particle is at O

Such vibration is called linearly polarized vibration.

(ii) δ = π/2 ⇒  which is equation of ellipse as shown in figure .The particle

moves in an elliptical path. The direction of its motion can be obtained as

x = A₁ cosωt and y = A₂ sinωt

At 1 t = 0, (ωt = 0) : x = A₁ , y = 0 i.e. particle is at M

At t = : x = 0, y = +A₂ i.e. particle is at P

At t = : x = −A₁ y = 0 i.e. particle is at Q

At t = :x =0, y= −A₂ i.e. particle is at L

At t = T (ωt = 2π) : x = +A₁ , y = 0 i.e. particle is at M

Thus the particle traces out an ellipse in the anti-clockwise direction. Such vibration is

called LEFT-HANDED elliptically polarized vibration.

In addition, If A₁ = A₂ = A , the motion become circular ( x² + y² = A² ) with radius A .

(iii) δ  = π ⇒

or  = 0 y =

This represents a equation of straight line, having negative slope -A₂/A₁ & passing through the origin.

(iv) δ  = 3π/2 =

Such vibration is called RIGHT-HANDED elliptically polarized vibration.

Addition of Two SHM having Different Frequencies (Lissajous Figures)

Frequencies in the Ratio of 2:1

Let a particle is subjected to two mutually perpendicular SHM having frequencies

ω₁ :ω₂ = 2 :1

x = A₁sin (2ωt +δ) and y = A₂ sinωt

Where A₁ & A₂ are the amplitude of the x -vibrations & y -vibration and δ is the phase

difference between them.

The equation of the curve of resultant motion is obtained as

x/A₁ = sin2ωt cosδ + cos2ωt sinδ = 2sinωt cosωt cosδ + (1 - 2sin²ωt)sinδ

=

⇒ .

⇒

⇒

Squaring and re-arranging terms, we get

 = 0

This is the general equation of curve having two loops

(i) Whenδ = 0 , sinδ = 0 , thus

This represents a curve symmetrical about both the axes.

(ii) When δ = π/2, sinδ = 1

= 0

This represents two coincident parabolas symmetrical about the x-axis and their vertices

at (A₁ ,0) . The equation of Parabola is

(iii) when δ =π , sinδ = 0 . The case is similar to case (i).

(iv) when  δ =3π/2 , sinδ = -1

so,

This represent parabola symmetric about the x -axis with their vertices at(-A₁ ,0) .

Following figures shows the Lissajous figures for various initial phase differences for

frequency ratio 2:1

Note: If particles displacements are of following form

x = A₁sin(2ωt) and y = A₂sin(ωt + δ)

The resultant Lissajous figures loops will be

(2) Frequencies in the ratio of 3:1

For x = A₁sin (3ωt +δ) and y = A₂sinωt

The resultant Lissajous figures at various initial phase differences are

Method to find the frequencies ratio:

To find the frequencies ratio from the given Lissajous figure, draw two lines parallel to x

and y axis which having maximum intercept with loops.

In above figure,  p_y = 6 & p_x = 2 ⇒ω₁ :ω₂ = 3:1