Superposition of Harmonic Oscillator-1.docx

Prof. Vilas shamrao patil

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,
�
2
??????
��
2
=−??????
2
??????
where the force(
??????
??????
??????
????????????
??????
)= 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,
�
2
??????
��
2
=−??????
2
??????
Where in the restoring force depends only on the displacement (y)
and does not contain higher order like y
2
, y
3
,---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 (??????⃗) is the vector sum of the individual
displacement that they can produce i.e. (??????⃗)= ((??????
1⃗⃗⃗⃗⃗)+(??????
2⃗⃗⃗⃗⃗) 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.
�
2
??????
��
2
=�−??????
2
??????+�??????
2
+�??????
3
+�??????
4
…….2
If y1 and y2 are two differential solutions of the differential equation.2
then we have

3. �
2
??????
1
��
2
=�−??????
2
??????
1+�??????
1
2
+�??????
1
3
+�??????
1
4
--------3
And
�
2
??????
2
��
2
=�−??????
2
??????
2+�??????
2
2
+�??????
2
3
+�??????
2
4
----------4
If the principle of superposition is true then y=y1+y2
Putting this in equation 2 becomes
�
2
??????
��
2
=
�
2
(??????
1+??????
2)
��
2
=�−??????
2
(??????
1+??????
2)+�(??????
1+??????
2)
2
+�(??????
1+??????
2)
3
+
�(??????
1+??????
2)
4
------5
Now adding equations 3 and 4 we get
�
2
??????
1
��
2
+
�
2
??????
2
��
2
=2�−??????
2
(??????
1+??????
2)+�(??????
1
2
+??????
2
2
)+�(??????
1
3
+??????
2
3
)
+�(??????
1
4
+??????
2
4
)
i.e.
�
2
??????
1
��
2
+
�
2
??????
2
��
2
=2�−??????
2
(??????
1+??????
2)+�(??????
1+??????
2)
2
+�(??????
1+??????
2)
3
+�(??????
1+??????
2)
4
−−6
Equations 5 and 6 are identical if
(
�
2
(??????
1+??????
2)
��
2
)=
�
2
??????
1
��
2
+
�
2
??????
2
��
2
-------7
A=2A----------8
−??????
2
(??????
1+??????
2)=−??????
2
??????
1−??????
2
??????
2−−−−−−−−9

4. �(??????
1+??????
2)
2
= �(??????
1
2
+??????
2
2
)−−−−10
�(??????
1+??????
2)
2
=�(??????
1
2
+??????
2
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 (??????) and different amplitudes (�
1,�
2)and different initial
phases (epoch angles) (∝
1,∝
2) as, ??????
1=�
1�??????�(??????�+∝
1)
…………1 and ??????
2=�
2�??????�(??????�+
∝
2) ………….2 where ??????
1 and
??????
2are the displacements of the particles that individuals simple
harmonics motions can produce.
∴ According to the principle of superposition,
y=??????
1+??????
2………….3
y=�
1�??????�(??????�+∝
1) +�
2�??????�(??????�+∝
2)------------
--4 using identity of trigonometry Sin
(A+B)=Sin A Cos B+ Cos A Sin B we get y=�
1(�??????�??????� ���∝
1+
���??????� �??????�∝
1)+�
2(�??????�??????� ���∝
2+���??????� �??????�∝
2) y=
(�
1���∝
1+�
2���∝
2)�??????�??????�+ (�
1�??????�∝
1+�
2�??????�∝
2) ���??????� -------5
since �
1,�
2 and ∝
1,∝
2 are the constants we can put
A cos∅ = �
1���∝
1+�
2���∝
2 --------6
A sin∅ = �
1�??????�∝
1+�
2�??????�∝
2 -------
--7 Now squaring

5. A
2
cos ∅
2
=
(�
1���∝
1+�
2���∝
2)
2

=�
1
2
���∝
1
2
+2�
1�
2���∝
1���∝
2 +�
2
2
���∝
2
2
------8
A
2
sin ∅
2
=(�
1�??????�∝
1+�
2�??????�∝
2)
2

=�
1
2
�??????�∝
1
2
+2�
1�
2�??????�∝
1�??????�∝
2
+�
2
2
�??????�∝
2
2
--------9 squaring and adding above two
equations 8 and 9
A
2
(sin ∅
2
+cos ∅
2
)=�
1
2
(�??????�∝
1
2
+���∝
1
2
) +
�
2
2
(�??????�∝
2
2
+���∝
2
2
)+2�
1�
2(���∝
1���∝
2+�??????�∝
1�??????�∝
2)--------
-------10
But in above equation,
(���∝
1���∝
2+�??????�∝
1�??????�∝
2)=���(∝
1−∝
2) and
using identity (sin ∅
2
+cos ∅
2
) = 1 above equation is
A
2
=
�
1
2
+�
2
2
+2�
1�
2 ���(∝
1−∝
2)---------------------11
and taking ratio of equation 6 and 7

A sin∅
A cos∅
=tan∅=
??????
1�??????�∝
1+??????
2�??????�∝
2
??????
1���∝
1+??????
2���∝
2
---------- ------12
in terms of A, ∅ the equation 3 becomes
y= � ��� ∅ �??????�??????�+� �??????� ∅ ���??????�
y=A sin (??????�+∅)-------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= �
1+�
2 ��� ∅=∝

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 �
1=OP and �
2=OQ are the
amplitudes of oscillations and ∝
1,∝
2are the initial phases of two SHM’s
with same frequency (??????), then at any instant (t) ��⃗⃗⃗⃗⃗⃗ and ��⃗⃗⃗⃗⃗⃗⃗ represent
radius vector as shown in fig. Therefore, the diagonal ��⃗⃗⃗⃗⃗⃗ 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 ∆���,∠��??????=∝
1= ∠��� using rule of parallel
line BP∥OV. Then sin∝
1=
��
��
=
??????
1
??????
1
then y1= �
1�??????�∝
1 and y2= �
2�??????�∝
2------
--------2
using 2 in 1 we get,
y=y1+y2=�
1�??????�∝
1+�
2�??????�∝
2---------------3
similarly considering projection on X-axis
RD=??????=??????
1+ ??????
2--------------4
In right angle triangle ∆�??????�,∠�??????�=∝
1.
Then Cos ∝
1=
�??????
��
=
??????
1
??????
1
then ??????
1= �
1���∝
1 and ??????
2= �
2���∝
2-----5 putting
in 4we get
RD=??????=??????
1+ ??????
2=�
1���∝
1+�
2���∝
2------6
In right angle triangle ∆��� , ,∠���=∅
Then tan ∅=
��
��
=
??????
1�??????�∝
1+??????
2�??????�∝
2
??????
1���∝
1+??????
2���∝
2
-----------7
Also from parallelogram law of vectors we know
OR
2
=OP
2
+OQ
2
+2OP.OQ.Cos (∠���)
i.e. OR
2
=A
2
=�
1
2
+�
2
2
+2�
1�
2���(∝
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 (??????�+∅) where (??????�+∅)is total phase angle. Similarly,
y1= OB = �
1�??????�(??????�+∝
1) and
y2=��=�
2�??????�(??????�+∝
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= &#3627408462; &#3627408480;??????&#3627408475;??????
1&#3627408481;=&#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481; and
y2= &#3627408463; &#3627408480;??????&#3627408475;??????
2&#3627408481;=&#3627408463; &#3627408480;??????&#3627408475; 2??????&#3627408475;
2&#3627408481;
The resultant displacement, according to superposition principle, is
y=y1+y2=&#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;+&#3627408463; &#3627408480;??????&#3627408475; 2??????&#3627408475;
2&#3627408481;
= &#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;+&#3627408463; &#3627408480;??????&#3627408475; 2??????[&#3627408475;
1−(&#3627408475;
1−&#3627408475;
2)]&#3627408481; - and + the same
quantity.
= &#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;+&#3627408463;&#3627408480;??????&#3627408475; [ 2??????&#3627408475;
1− 2??????(&#3627408475;
1−&#3627408475;
2)]&#3627408481;
=&#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;
+&#3627408463; [ &#3627408480;??????&#3627408475;2??????&#3627408475;
1&#3627408481;.&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;− &#3627408464;&#3627408476;&#3627408480;2??????&#3627408475;
1&#3627408481;.&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1
−&#3627408475;
2)&#3627408481;]
=&#3627408462; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;+(&#3627408463; &#3627408480;??????&#3627408475;2??????&#3627408475;
1&#3627408481;.&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;
−&#3627408463;&#3627408464;&#3627408476;&#3627408480;2??????&#3627408475;
1&#3627408481;.&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)
Taking common
= &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;(&#3627408462;+b&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)−&#3627408464;&#3627408476;&#3627408480;2??????&#3627408475;
1&#3627408481;.(&#3627408463;&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1−
&#3627408475;
2)&#3627408481;)
Put (&#3627408462;+b&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)= A.cos??????and
Put (&#3627408463;&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)=A. sin ??????
Putting we get ??????=&#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;.A.cos??????−&#3627408464;&#3627408476;&#3627408480;2??????&#3627408475;
1&#3627408481;.Asin ??????
Squaring and adding above equations,

10. A
2
(sin
2
??????+&#3627408464;&#3627408476;&#3627408480;
2
??????)= (&#3627408462;+b&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)
2
+(&#3627408463;&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1−
&#3627408475;
2)&#3627408481;)
2

A
2
=a
2
+2ab&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+b
2
&#3627408464;&#3627408476;&#3627408480;
2
2??????(&#3627408475;
1−&#3627408475;
2)t +&#3627408463;
2
&#3627408480;??????&#3627408475;
2
2??????(&#3627408475;
1−
&#3627408475;
2)&#3627408481;
As using identity (sin
2
??????+&#3627408464;&#3627408476;&#3627408480;
2
??????)=1
A
2
= a
2
+ 2ab&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+b
2
(&#3627408480;??????&#3627408475;
2
2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+&#3627408464;&#3627408476;&#3627408480;
2
2??????(&#3627408475;
1−
&#3627408475;
2)t)
A
2
= a
2
+ 2ab&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+b
2

A=(a
2
+ 2ab&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+ b
2
)
1
2
And tan ??????=
&#3627408454;??????&#3627408475;??????
&#3627408438;&#3627408476;&#3627408480;??????
=
(&#3627408463;&#3627408480;??????&#3627408475; 2??????(&#3627408475;1−&#3627408475;
2)
&#3627408481;)
??????
(&#3627408462;+b&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;1−&#3627408475;
2)
&#3627408481;)
??????
=
(b&#3627408480;??????&#3627408475; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)
(??????+b&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;)

y= &#3627408436; &#3627408480;??????&#3627408475; 2??????&#3627408475;
1&#3627408481;.&#3627408464;&#3627408476;&#3627408480; ?????? −&#3627408436; &#3627408464;&#3627408476;&#3627408480; 2??????&#3627408475;
2&#3627408481; . &#3627408480;??????&#3627408475;??????
y= &#3627408436; &#3627408480;??????&#3627408475; (2??????&#3627408475;
1&#3627408481;−??????)--------------10
is resultant displacement with amplitude
A=√a
2
+ 2ab&#3627408464;&#3627408476;&#3627408480; 2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;+ b
2

I)Maxima- For maximum sound the amplitude must be maximum at
certain time t, which requires the condition
2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;=2??????.&#3627408472; where k=1,2,3,……
i.e. at time, t=
??????
(&#3627408475;
1−&#3627408475;
2)
i.e. at instances 0,
1
(&#3627408475;
1−&#3627408475;
2)
,
2
(&#3627408475;
1−&#3627408475;
2)
,
3
(&#3627408475;
1−&#3627408475;
2)
,------

11. The amplitude and hence sound (loudness) is maximum.
Beat period =
1
(&#3627408475;
1−&#3627408475;
2)

Beat frequency, N= (&#3627408475;
1−&#3627408475;
2)
(II)Minima-For minimum loudness the amplitude (A) must be minimum
which requires the condition that,
2??????(&#3627408475;
1−&#3627408475;
2)&#3627408481;=(2&#3627408472;+1)?????? where k=0,1,2,3------
??????.&#3627408466;. at time &#3627408481;=
2??????+1
2(&#3627408475;
1−&#3627408475;
2),
,
&#3627408462;&#3627408481; &#3627408481;??????&#3627408474;&#3627408466; instance
1
2(&#3627408475;
1−&#3627408475;
2)
,
3
2(&#3627408475;
1−&#3627408475;
2)
,
5
2(&#3627408475;
1−&#3627408475;
2)
,------
&#3627408437;&#3627408466;&#3627408462;&#3627408481; periods, T=
1
2(&#3627408475;
1−&#3627408475;
2)

&#3627408437;&#3627408466;&#3627408462;&#3627408481; ??????&#3627408479;&#3627408466;&#3627408478;&#3627408482;&#3627408466;&#3627408475;&#3627408464;??????,&#3627408449;=
1
&#3627408455;
=2(&#3627408475;
1−&#3627408475;
2),
??????&#3627408481; &#3627408480;ℎ&#3627408476;&#3627408482;&#3627408473;&#3627408465; &#3627408463;&#3627408466; &#3627408475;&#3627408476;&#3627408481;&#3627408466;&#3627408465; &#3627408481;ℎ&#3627408462;&#3627408481; if a=b, then the minimum amplitude becomes
zero and hence maxima and minima are more distinct. Hence beats are
heard very clearly.