1) When a particle undergoes two simple harmonic oscillations simultaneously, the resultant displacement is the sum of the individual displacements according to the principle of superposition. 2) Only linear homogeneous differential equations obey the principle of superposition. For nonlinear or non-homogeneous equations, the sum of the solutions is not a solution. 3) The superposition of two collinear harmonic oscillations of equal frequency results in a simple harmonic motion. The amplitude and phase of the resultant oscillation can be determined using analytical or graphical methods.

1. Superposition of Harmonic Oscillations
1.1 : Introduction We know that a linear simple harmonic motion
(SHM) is represented by a homogeneous linear differential equation,
𝑑2
𝑦
𝑑𝑡2
= −𝑤2
𝑦
where the force(
𝒅𝟐𝒚
𝒅𝒕𝟐 )= acting on the particle is proportional to the
displacement (y) from mean position and is always directed towards the
mean position, which is shown by the negative sign. Important property
of such homogeneous linear equations is that the sum of any two
solutions is also a solution.
When a particle is subjected to two simple harmonic oscillations
simultaneously, then the resultant displacement is given by
y = y1+y2
where y1and y2 are displacements produced by individual simple
harmonic motions.
This is known as principle of superposition which is possible only
in case of homogeneous linear equations.
The resultant motion of the particle traces a curve called Lissajous
figure.
The shape of Lissajous figure depends on the frequency, amplitude
and phase difference of the two-constituent simple harmonic
oscillations. The study of Lissajous figures is useful in comparing the
frequencies.
1.2 : Linearity and Superposition Principle

2. A simple harmonic motion is generally represented by a differential
equation given by,
𝑑2
𝑦
𝑑𝑡2
= −𝑤2
𝑦
Where in the restoring force depends only on the displacement (y)
and does not contain higher order like y2, y3,---etc, Such equation are
called linear equations.If equations contain higher order term then it is
called Non- Linear equation.
Secondly, we noticed that equation 1 does not contain terms
independent of y and hence it is called as homogeneous linear
equation. If equation contains terms independent of (y) then it is called
non-homogeneous equation.
Important property of homogeneous, linear equation is that the
sum of two solution equation 1 is homogeneous linear.
The principle of superposition states that when a particle is
subjected to two simple harmonic motion simultaneously then the
resultant displacement (𝑦
⃗) is the vector sum of the individual
displacement that they can produce i.e. (𝑦
⃗)= ((𝑦1
⃗⃗⃗⃗⃗) + (𝑦2
⃗⃗⃗⃗⃗) It is important
to note that only homogeneous linear equations obey the principle of
superposition.
To prove above statement let us consider a non-homogeneous non-
linear differential equation.
𝑑2𝑦
𝑑𝑡2 = 𝐴 − 𝑤2
𝑦 + 𝐵𝑦2
+ 𝐶𝑦3
+ 𝐷𝑦4
… ….2
If y1 and y2 are two differential solutions of the differential equation.2
then we have

3. 𝑑2𝑦1
𝑑𝑡2 = 𝐴 − 𝑤2
𝑦1 + 𝐵𝑦1
2
+ 𝐶𝑦1
3
+ 𝐷𝑦1
4
--------3
And
𝑑2𝑦2
𝑑𝑡2 = 𝐴 − 𝑤2
𝑦2 + 𝐵𝑦2
2
+ 𝐶𝑦2
3
+ 𝐷𝑦2
4
----------4
If the principle of superposition is true then y=y1+y2
Putting this in equation 2 becomes
𝑑2𝑦
𝑑𝑡2 =
𝑑2(𝑦1+𝑦2)
𝑑𝑡2 = 𝐴 − 𝑤2
(𝑦1 + 𝑦2) + 𝐵(𝑦1 + 𝑦2)2
+ 𝐶(𝑦1 + 𝑦2)3
+
𝐷(𝑦1 + 𝑦2)4
------5
Now adding equations 3 and 4 we get
𝑑2
𝑦1
𝑑𝑡2
+
𝑑2
𝑦2
𝑑𝑡2
= 2𝐴 − 𝑤2(𝑦1 + 𝑦2) + 𝐵(𝑦1
2
+ 𝑦2
2) + 𝐶(𝑦1
3
+ 𝑦2
3)
+ 𝐷(𝑦1
4
+ 𝑦2
4
)
i.e.
𝑑2
𝑦1
𝑑𝑡2
+
𝑑2
𝑦2
𝑑𝑡2
= 2𝐴 − 𝑤2
(𝑦1 + 𝑦2) + 𝐵(𝑦1 + 𝑦2)2
+ 𝐶(𝑦1 + 𝑦2)3
+ 𝐷(𝑦1 + 𝑦2)4
− −6
Equations 5 and 6 are identical if
(
𝑑2(𝑦1+𝑦2)
𝑑𝑡2 ) =
𝑑2𝑦1
𝑑𝑡2 +
𝑑2𝑦2
𝑑𝑡2 -------7
A=2A----------8
−𝑤2(𝑦1 + 𝑦2) = −𝑤2
𝑦1 − 𝑤2
𝑦2 − − − − − − − −9

4. 𝐵(𝑦1 + 𝑦2)2
= 𝐵(𝑦1
2
+ 𝑦2
2) − − − −10
𝐶(𝑦1 + 𝑦2)2
= 𝐶(𝑦1
2
+ 𝑦2
2) − − − −11
It can be noted that equation 7 and 9 are true but equations
8,10,11 are true only if when constants A,B,C are identically zero(0).Thus
only homogeneous equations obey principle of superposition.
Superposition of Two Collinear Harmonic Oscillation having equal
frequency: -
(i)Analytical Method :- Consider two simple harmonic motions with same
frequency (𝜔) and different amplitudes (𝑎1,𝑎2)and different initial
phases (epoch angles) (∝1,∝2) as, 𝑦1=𝑎1𝑠𝑖𝑛(𝜔𝑡 +∝1)
…………1 and 𝑦2=𝑎2𝑠𝑖𝑛(𝜔𝑡 +
∝2) ………….2 where 𝑦1 and
𝑦2are the displacements of the particles that individuals simple
harmonics motions can produce.
∴ According to the principle of superposition,
y=𝑦1+𝑦2………….3
y=𝑎1𝑠𝑖𝑛(𝜔𝑡 +∝1) +𝑎2𝑠𝑖𝑛(𝜔𝑡 +∝2)------------
--4 using identity of trigonometry Sin
(A+B)=Sin A Cos B+ Cos A Sin B we get y=𝑎1(𝑠𝑖𝑛𝜔𝑡 𝑐𝑜𝑠 ∝1+
𝑐𝑜𝑠𝜔𝑡 𝑠𝑖𝑛 ∝1)+𝑎2(𝑠𝑖𝑛𝜔𝑡 𝑐𝑜𝑠 ∝2+ 𝑐𝑜𝑠𝜔𝑡 𝑠𝑖𝑛 ∝2) y=
(𝑎1𝑐𝑜𝑠 ∝1+ 𝑎2𝑐𝑜𝑠 ∝2)𝑠𝑖𝑛𝜔𝑡+ (𝑎1𝑠𝑖𝑛 ∝1+ 𝑎2𝑠𝑖𝑛 ∝2) 𝑐𝑜𝑠𝜔𝑡 -------5
since 𝑎1,𝑎2 and ∝1,∝2 are the constants we can put
A cos∅ = 𝑎1𝑐𝑜𝑠 ∝1+ 𝑎2𝑐𝑜𝑠 ∝2 --------6
A sin∅ = 𝑎1𝑠𝑖𝑛 ∝1+ 𝑎2𝑠𝑖𝑛 ∝2 -------
--7 Now squaring

5. A2
cos ∅2
=
(𝑎1𝑐𝑜𝑠 ∝1+ 𝑎2𝑐𝑜𝑠 ∝2)2
=𝑎1
2
𝑐𝑜𝑠 ∝1
2
+2𝑎1𝑎2𝑐𝑜𝑠 ∝1 𝑐𝑜𝑠 ∝2 +𝑎2
2
𝑐𝑜𝑠 ∝2
2
------8
A2
sin ∅2
=(𝑎1𝑠𝑖𝑛 ∝1+ 𝑎2𝑠𝑖𝑛 ∝2)2
=𝑎1
2
𝑠𝑖𝑛 ∝1
2
+2𝑎1𝑎2𝑠𝑖𝑛 ∝1 𝑠𝑖𝑛 ∝2
+𝑎2
2
𝑠𝑖𝑛 ∝2
2
--------9 squaring and adding above two
equations 8 and 9
A2
(sin ∅2
+cos ∅2
)=𝑎1
2
(𝑠𝑖𝑛 ∝1
2
+ 𝑐𝑜𝑠 ∝1
2
) +
𝑎2
2
(𝑠𝑖𝑛 ∝2
2
+ 𝑐𝑜𝑠 ∝2
2
)+2𝑎1𝑎2(𝑐𝑜𝑠 ∝1 𝑐𝑜𝑠 ∝2+ 𝑠𝑖𝑛 ∝1 𝑠𝑖𝑛 ∝2)--------
-------10
But in above equation,
(𝑐𝑜𝑠 ∝1 𝑐𝑜𝑠 ∝2+ 𝑠𝑖𝑛 ∝1 𝑠𝑖𝑛 ∝2) = 𝑐𝑜𝑠(∝1−∝2) and
using identity (sin ∅2
+cos ∅2
) = 1 above equation is
A2
=
𝑎1
2
+𝑎2
2
+2𝑎1𝑎2 𝑐𝑜𝑠(∝1−∝2)---------------------11
and taking ratio of equation 6 and 7
A sin∅
A cos∅
=tan∅ =
𝑎1𝑠𝑖𝑛∝1+𝑎2𝑠𝑖𝑛∝2
𝑎1𝑐𝑜𝑠∝1+𝑎2𝑐𝑜𝑠∝2
---------- ------12
in terms of A, ∅ the equation 3 becomes
y= 𝑎 𝑐𝑜𝑠 ∅ 𝑠𝑖𝑛𝜔𝑡 + 𝑎 𝑠𝑖𝑛 ∅ 𝑐𝑜𝑠𝜔𝑡
y=A sin (𝜔𝑡 + ∅)-------13 This equation 13 represent the
resultant SHM whose amplitude A is given by equation 11 and epoch(∅)
is given by equation 12 in special case if ∝1=∝2 =∝
then A= 𝑎1 + 𝑎2 𝑎𝑛𝑑 ∅ =∝

6. Graphical Method
Let OB=y1 and OC=y2 be the displacement about Y-axis due to two
different SHM’s at any instant of time (t). If 𝑎1=OP and 𝑎2=OQ are the
amplitudes of oscillations and ∝1, ∝2are the initial phases of two SHM’s
with same frequency (𝜔), then at any instant (t) 𝑂𝑃
⃗⃗⃗⃗⃗⃗ and 𝑂𝑄
⃗⃗⃗⃗⃗⃗⃗ represent
radius vector as shown in fig. Therefore, the diagonal 𝑂𝑅
⃗⃗⃗⃗⃗⃗ of the
parallelogram OPRQ is the resultant radius vector and hence resultant
displacement along Y-axis is OD=y.
Since the projection OP on y-axis (i.e.OB) =projection of QR on Y -axis
(i.e.CD), then resultant displacement along Y-axis is given by

7. y=OD=OC+CD=OC+OB=y2+y1=y1+y2---------1
In right angle triangle ∆𝑂𝑃𝐵, ∠𝑃𝑂𝑉 =∝1= ∠𝑃𝑂𝐵 using rule of parallel
line BP∥OV. Then sin∝1=
𝑂𝐵
𝑂𝑃
=
𝑦1
𝑎1
then y1= 𝑎1𝑠𝑖𝑛 ∝1 and y2= 𝑎2𝑠𝑖𝑛 ∝2------
--------2
using 2 in 1 we get,
y=y1+y2=𝑎1𝑠𝑖𝑛 ∝1+ 𝑎2𝑠𝑖𝑛 ∝2---------------3
similarly considering projection on X-axis
RD=𝜘=𝜘1+ 𝜘2--------------4
In right angle triangle ∆𝑂𝑉𝑃, ∠𝑂𝑉𝑃 =∝1.
Then Cos ∝1=
𝑂𝑉
𝑂𝑃
=
𝜘1
𝑎1
then 𝜘1= 𝑎1𝐶𝑜𝑠 ∝1 and 𝜘2= 𝑎2𝐶𝑜𝑠 ∝2-----5 putting
in 4we get
RD=𝜘=𝜘1+ 𝜘2 = 𝑎1𝐶𝑜𝑠 ∝1+𝑎2𝐶𝑜𝑠 ∝2------6
In right angle triangle ∆𝑂𝑅𝐷 , , ∠𝑂𝑅𝐷 = ∅
Then tan ∅=
𝑂𝐷
𝑅𝐷
=
𝑎1𝑠𝑖𝑛∝1+𝑎2𝑠𝑖𝑛∝2
𝑎1𝐶𝑜𝑠∝1+𝑎2𝐶𝑜𝑠∝2
-----------7
Also from parallelogram law of vectors we know
OR2=OP2+OQ2+2OP.OQ.Cos (∠𝑄𝑂𝑃)
i.e. OR2=A2=𝑎1
2
+𝑎2
2
+2𝑎1𝑎2𝐶𝑜𝑠(∝1 −∝2 )-----------------8
This satisfy the amplitude relation.
Thus, diagonal OR represents completely the resultant of two collinear
SHM's. The resultant amplitude (A) and epoch ∅ are given by the
equations (8) and (7) respectively.

8. The resultant displacement of the particle along y-axis is given by,
y =Asin (𝜔𝑡 + ∅) where (𝜔𝑡 + ∅)is total phase angle. Similarly,
y1= OB = 𝑎1𝑠𝑖𝑛(𝜔𝑡 +∝1) and
y2=𝑂𝐶 = 𝑎2𝑠𝑖𝑛(𝜔𝑡 +∝2)
1:4. Superposition of Two Collinear Harmonic Oscillations Having
Different Frequencies (Beats)
If two sounding sources like tuning forks or musical instruments of
nearly same frequency and amplitude are sounded together, then at any
given point the phase difference between two wave trains meeting goes
on changing continuously. Thus, at sometime the waves meet in phase
when maximum sound is produced and next time when they meet out
of phase minimum sound is produced. Thus, alternately maxima and
minima of sound are produced which is called waxing and waning of
sound and are known as beats. Beats are clearly heard of the beat
frequency (i.e. number of beats heard per second) is less than ten, which
is due to perception of ear. Measurement of beat frequency helps in
determining the frequency of sounding sources.
Analytical Treatment of Beats
Consider two sounding sources with frequencies n1 and n2 such that (n1~
n2) < 10. Let a and b be the amplitudes of the waves respectively. Let us
suppose that at time t = 0, the two waves are in phase at a point in the
medium.

9. Therefore, individual displacements produced by the two waves are
given by,
y1= 𝑎 𝑠𝑖𝑛𝜔1𝑡 = 𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡 and
y2= 𝑏 𝑠𝑖𝑛𝜔2𝑡 = 𝑏 𝑠𝑖𝑛 2𝜋𝑛2𝑡
The resultant displacement, according to superposition principle, is
y=y1+y2=𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡 + 𝑏 𝑠𝑖𝑛 2𝜋𝑛2𝑡
= 𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡 + 𝑏 𝑠𝑖𝑛 2𝜋[𝑛1 − (𝑛1 − 𝑛2)]𝑡 - and + the same
quantity.
= 𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡 + 𝑏𝑠𝑖𝑛 [ 2𝜋𝑛1 − 2𝜋(𝑛1 − 𝑛2)]𝑡
= 𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡
+ 𝑏 [ 𝑠𝑖𝑛2𝜋𝑛1𝑡. 𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡 − 𝑐𝑜𝑠2𝜋𝑛1𝑡. 𝑠𝑖𝑛 2𝜋(𝑛1
− 𝑛2)𝑡]
= 𝑎 𝑠𝑖𝑛 2𝜋𝑛1𝑡 + (𝑏 𝑠𝑖𝑛2𝜋𝑛1𝑡. 𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡
− 𝑏𝑐𝑜𝑠2𝜋𝑛1𝑡. 𝑠𝑖𝑛 2𝜋(𝑛1 − 𝑛2)𝑡)
Taking common
= 𝑠𝑖𝑛 2𝜋𝑛1𝑡(𝑎 + b𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡) − 𝑐𝑜𝑠2𝜋𝑛1𝑡. (𝑏𝑠𝑖𝑛 2𝜋(𝑛1 −
𝑛2)𝑡)
Put (𝑎 + b𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡)= A.cos𝜃and
Put (𝑏𝑠𝑖𝑛 2𝜋(𝑛1 − 𝑛2)𝑡)=A. sin 𝜃
Putting we get 𝑦 = 𝑠𝑖𝑛 2𝜋𝑛1𝑡. A. cos𝜃 − 𝑐𝑜𝑠2𝜋𝑛1𝑡. Asin 𝜃
Squaring and adding above equations,

10. A2(sin2 𝜃 + 𝑐𝑜𝑠2
𝜃) = (𝑎 + b𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡)2
+ (𝑏𝑠𝑖𝑛 2𝜋(𝑛1 −
𝑛2)𝑡)2
A2=a2+2ab𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡+b2
𝑐𝑜𝑠2
2𝜋(𝑛1 − 𝑛2)t +𝑏2
𝑠𝑖𝑛2
2𝜋(𝑛1 −
𝑛2)𝑡
As using identity (sin2 𝜃 + 𝑐𝑜𝑠2
𝜃)=1
A2= a2 + 2ab𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡+b2
(𝑠𝑖𝑛2
2𝜋(𝑛1 − 𝑛2)𝑡 + 𝑐𝑜𝑠2
2𝜋(𝑛1 −
𝑛2)t)
A2= a2 + 2ab𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡+b2
A=(a2
+ 2ab𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡 + b2
)
1
2
And tan 𝜃 =
𝑆𝑖𝑛𝜃
𝐶𝑜𝑠𝜃
=
(𝑏𝑠𝑖𝑛 2𝜋(𝑛1−𝑛2)𝑡)
𝐴
(𝑎+b𝑐𝑜𝑠 2𝜋(𝑛1−𝑛2)𝑡)
𝐴
=
(b𝑠𝑖𝑛 2𝜋(𝑛1−𝑛2)𝑡)
(𝑎+b𝑐𝑜𝑠 2𝜋(𝑛1−𝑛2)𝑡)
y= 𝐴 𝑠𝑖𝑛 2𝜋𝑛1𝑡. 𝑐𝑜𝑠 𝜃 − 𝐴 𝑐𝑜𝑠 2𝜋𝑛2𝑡 . 𝑠𝑖𝑛𝜃
y= 𝐴 𝑠𝑖𝑛 (2𝜋𝑛1𝑡 − 𝜃)--------------10
is resultant displacement with amplitude
A=√a2 + 2ab𝑐𝑜𝑠 2𝜋(𝑛1 − 𝑛2)𝑡 + b2
I)Maxima- For maximum sound the amplitude must be maximum at
certain time t, which requires the condition
2𝜋(𝑛1 − 𝑛2)𝑡 = 2𝜋. 𝑘 where k=1,2,3,……
i.e. at time, t=
𝑘
(𝑛1−𝑛2)
i.e. at instances 0,
1
(𝑛1−𝑛2)
,
2
(𝑛1−𝑛2)
,
3
(𝑛1−𝑛2)
,------

11. The amplitude and hence sound (loudness) is maximum.
Beat period =
1
(𝑛1−𝑛2)
Beat frequency, N= (𝑛1 − 𝑛2)
(II)Minima-For minimum loudness the amplitude (A) must be minimum
which requires the condition that,
2𝜋(𝑛1 − 𝑛2)𝑡 = (2𝑘 + 1)𝜋 where k=0,1,2,3------
𝑖. 𝑒. at time 𝑡 =
2𝑘+1
2(𝑛1−𝑛2),
,
𝑎𝑡 𝑡𝑖𝑚𝑒 instance
1
2(𝑛1−𝑛2)
,
3
2(𝑛1−𝑛2)
,
5
2(𝑛1−𝑛2)
,------
𝐵𝑒𝑎𝑡 periods, T=
1
2(𝑛1−𝑛2)
𝐵𝑒𝑎𝑡 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦, 𝑁 =
1
𝑇
= 2(𝑛1 − 𝑛2),
𝐼𝑡 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑛𝑜𝑡𝑒𝑑 𝑡ℎ𝑎𝑡 if a=b, then the minimum amplitude becomes
zero and hence maxima and minima are more distinct. Hence beats are
heard very clearly.