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Composition of harmonic motions

Module by: Sunil Kumar Singh. E-mail the author

Composition, here, means combining more than one simple harmonic motion. However, this statement needs to be interpreted carefully. Every particle has only one motion at a time. Actually, we mean to combine two or more harmonic motions, which result from the operation of forces, each of which is individually capable of producing SHM. Therefore, it is actually combining SHMs, produced by forces operating on a “single” particle.

Further, we need to emphasize one important limitation to our discussion. Our context is limited to combining effects of forces which produce SHMs of “same” frequency.

We organize our study under two heads :

  • Composition of SHMs along same straight line.
  • Composition of SHMs along two mutually perpendicular straight lines.

Force analysis

The motion of a particle, acted upon by two or more forces, is governed by Newton’s second law. The situation, here, is no different except that each of the forces is characterized for being proportional to negative of displacement. This means that each of the forces, if left to act alone, would produce SHM. Nonetheless, we have seen that force is a vector quantity with the underlying characteristic that each force produces its effect independent of other force.

The independence of a force to the presence of other forces makes our task easy to assess the net result. Let us consider the motion of a particle as if it is being worked alone by a single force. Let “ r 1 r 1 ” and “ v 1 v 1 ” be the position and velocity at a particular instant resulting from the action of this force. Now, let “ r 2 r 2 ” and “ v 2 v 2 ” be the position and velocity at a particular instant resulting from the action of the other force as if it is the only force working on the particle.

The resultant position and velocity vector, then, would simply be the vector additions of individual quantities. Hence, position and velocity of the particle, when operated simultaneously by two forces, would be :.

r = r 1 + r 2 r = r 1 + r 2

v = v 1 + v 2 v = v 1 + v 2

These equations provide the general framework for studying motion that result from action of more than one force capable of producing SHM.

Composition of SHMs along same straight line

Let us consider two SHM forces, F 1 F 1 and F 2 F 2 , acting along the same straight line. Let the displacements be given by two equations,

x 1 = A 1 sin ω t x 1 = A 1 sin ω t

x 2 = A 2 sin ω t + φ x 2 = A 2 sin ω t + φ

We have written two displacements which reflect a convenient general case. Amplitudes are different. At any given instant, one of the two SHMs is “ahead of” or “lags behind” other, depending on the sign of phase constant “φ”. As pointed out earlier, we have kept the angular frequency “ω” same for both SHMs.

Now, we want to find the net displacement of the particle at any given instant. Referring to our earlier discussion, we can find net displacement by evaluating vector relation,

r = r 1 + r 2 r = r 1 + r 2

Since both SHMs are along the same straight line, we can drop the vector sign and can simply write this relation in the present context as :

x = x 1 + x 2 x = x 1 + x 2

x = A 1 sin ω t + A 2 sin ω t + φ x = A 1 sin ω t + A 2 sin ω t + φ

Expanding trigonometric function,

x = A 1 sin ω t + A 2 sin ω t cos φ + A 2 cos ω t sin φ x = A 1 sin ω t + A 2 sin ω t cos φ + A 2 cos ω t sin φ

Segregating sine and cosine functions, keeping in mind that “φ” is constant :

x = A 1 + A 2 cos φ sin ω t + A 2 sin φ cos ω t x = A 1 + A 2 cos φ sin ω t + A 2 sin φ cos ω t

The expressions in the brackets are constant. Let,

C = A 1 + A 2 cos φ and D = A 2 sin φ C = A 1 + A 2 cos φ and D = A 2 sin φ

Substituting in the expression of displacement, we have :

x = C sin ω t + D cos ω t x = C sin ω t + D cos ω t

Following standard analytical method, Let

C = A cos θ and D = A sin θ C = A cos θ and D = A sin θ

Substituting in the expression of displacement again, we have :

x = A cos θ sin ω t + A sin θ cos ω t x = A cos θ sin ω t + A sin θ cos ω t

x = A sin ω t + θ x = A sin ω t + θ

This is the final expression of the composition of two SHMs in the same straight line. Clearly, the amplitude of resulting SHM is “A”. Also, the resulting SHM differs in phase with respect to either of the two SHMs. In particular, the phase of resulting SHM differs by an angle “θ” with respect of first SHM, whose displacement is given by “ A 1 sin ω t A 1 sin ω t ” We also note that frequency of the resulting SHM is same as either of two SHMs.

Phase constant

The phase constant of the resulting SHM is :

tan θ = D C = A 2 sin φ A 1 + A 2 cos φ tan θ = D C = A 2 sin φ A 1 + A 2 cos φ

Amplitude

The amplitude of the resultant harmonic motion is obtained solving substitutions made in the derivation.

C = A cos θ and D = A sin θ C = A cos θ and D = A sin θ

A = C 2 + D 2 = { A 1 + A 2 cos φ 2 + A 2 sin φ 2 } A = C 2 + D 2 = { A 1 + A 2 cos φ 2 + A 2 sin φ 2 }

A = A 1,2 + A 2 2 cos 2 φ + 2 A 1 A 2 cos φ + A 2 2 sin 2 φ A = A 1,2 + A 2 2 cos 2 φ + 2 A 1 A 2 cos φ + A 2 2 sin 2 φ

A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ

Important cases

We ,here, consider few interesting cases :

1: Phase difference is zero

Two SHMs are in same phase. In this case, cos φ = cos 0 0 = 1 cos φ = cos 0 0 = 1 .

A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ = A 1 2 + A 2 2 + 2 A 1 A 2 = A 1 + A 2 2 A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ = A 1 2 + A 2 2 + 2 A 1 A 2 = A 1 + A 2 2

A = A 1 + A 2 A = A 1 + A 2

If additionally A 1 = A 2 A 1 = A 2 , then A = 2 A 1 = 2 A 2 A = 2 A 1 = 2 A 2 . Further, phase constant is given by :

tan θ = D C = A 2 sin φ A 1 + A 2 cos φ = A 2 sin 0 0 A 1 + A 2 cos 0 0 = 0 tan θ = D C = A 2 sin φ A 1 + A 2 cos φ = A 2 sin 0 0 A 1 + A 2 cos 0 0 = 0

θ = 0 θ = 0

2: Phase difference is “π”

Two SHMs are opposite in phase. In this case, cos φ = cos π = - 1 cos φ = cos π = - 1 .

A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ = A 1 2 + A 2 2 2 A 1 A 2 = A 1 A 2 2 A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ = A 1 2 + A 2 2 2 A 1 A 2 = A 1 A 2 2

A = A 1 A 2 A = A 1 A 2

The amplitude is a non-negative number. In order to reflect this aspect, we write amplitude in modulus form :

A = | A 1 A 2 | A = | A 1 A 2 |

If additionally A 1 = A 2 A 1 = A 2 , then A = 0. In this case, the particle does not oscillate. Further, phase constant is given by :

tan θ = D C = A 2 sin φ A 1 + A 2 cos φ = A 2 sin π A 1 + A 2 cos π = 0 tan θ = D C = A 2 sin φ A 1 + A 2 cos φ = A 2 sin π A 1 + A 2 cos π = 0

θ = 0 θ = 0

Composition by vector method

We have evaluated composition of two SHMs analytically. This has given us the detailed picture of how displacement of a particle takes place. In the nutshell, we find that resulting motion is also a SHM of same frequency as that of constituting SHMs. Besides, we are able to determine followings aspects of resulting SHM :

  • Displacement
  • Amplitude of the resulting SHM
  • Phase constant of the resulting SHM

There is, however, an effective and more convenient alternative to determine all these aspects of SHM, using vector concept. The important thing to realize here is that amplitude can be associated with direction – apart from having magnitude. Its direction is qualified by the phase constant.

The equation of amplitude, derived earlier, provides the basis of this assumption. If we look closely at the expression of resultant amplitude, then we realize that the expression actually represents sum of two vectors namely “ A 1 A 1 ” and “ A 2 A 2 ” at an angle “φ” as shown in the figure.

Figure 1: The diagonal represents the resultant amplitude.
Composition of two SHMs
 Composition of two SHMs  (cs1.gif)

A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ A = A 1 2 + A 2 2 + 2 A 1 A 2 cos φ

This understanding serves our purpose. If we know amplitudes of individual SHMs and the phase difference, then we can find amplitude of the resulting SHM directly using vector sum formula. It is also evident that vector method can be used to find the resulting phase difference. From the figure,

tan θ = CD OD = A 2 sin φ A 1 + A 2 cos φ tan θ = CD OD = A 2 sin φ A 1 + A 2 cos φ

Vector method has one more simplifying aspect. We can compose more than two SHMs by consecutive application of parallelogram theorem or by using consecutive application of triangle law of vector addition. See the figure and observe how do we compose three SHMs and find the resulting amplitude (A) and phase difference (θ) from the reference direction.

Figure 2: The closing side of the ploygon represents the resultant amplitude.
Composition of more than two SHMs
 Composition of more than two SHMs  (cs2.gif)

Composition of SHMs in perpendicular directions

Simple harmonic motions in mutually perpendicular directions constitute a two dimensional motion. The x-direction of motion is governed by SHM force in x-direction, whereas y-direction of motion is governed by SHM force in y-direction. Let the displacements in “x” and “y” directions be :

x = A sin ω t x = A sin ω t

y = B sin ω t + φ y = B sin ω t + φ

The force in x-direction does not affect the displacement in y-direction and vice versa. We have pointed out that “effect” of one force is independent of the presence of other forces. Here, this independence is a step ahead. Force in one direction is incapable to produce “effect” in perpendicular direction in the first place. As such, the two equations, as they are, give “x” and “y” coordinates of the particle at any given instant. The resulting motion is two dimensional motion. In the nutshell, we do not need to combine effects (displacements) in a particular direction as in the case of one dimensional composition.

Lissajous curves

The plot of the motion as defined by the SHM displacement equations is known as “Lissajous curve”. In order to determine the curve (path) of resulting motion, we eliminate “t” from two equations. For this, we first expand the trigonometric expression of displacement in “y” direction :

y = B sin ω t cos φ + cos ω t sin φ y = B sin ω t cos φ + cos ω t sin φ

Now, we substitute the values of trigonometric functions from the expression of displacement in “x” direction. Here,

sin ω t = x A and cos ω t = 1 x 2 A 2 sin ω t = x A and cos ω t = 1 x 2 A 2

Substituting these values,

y = B { x A cos φ + 1 x 2 A 2 sin φ } y = B { x A cos φ + 1 x 2 A 2 sin φ }

y B - x A cos φ = 1 x 2 A 2 sin φ y B - x A cos φ = 1 x 2 A 2 sin φ

Squaring both sides and rearranging,

x 2 A 2 + y 2 B 2 2 x y cos φ A B = sin 2 φ x 2 A 2 + y 2 B 2 2 x y cos φ A B = sin 2 φ

This is an equation of ellipse. The nature of path depends on amplitudes of the individual SHMs and the phase difference. Importantly, we realize that resulting motion may not be oscillatory at all – although it is periodic.

Figure 3: The path of motion is an ellipse.
Lissajous curves
 Lissajous curves  (cs3.gif)

Further, we observe from the equation for displacement in “x” direction that values of “x” lie between “-A” and “A”. Similarly, values of “y” lie between “-B” and “B”. Clearly, path of resulting motion (i.e. ellipse) lies within the boundary, set up by limiting values of “x” and “y”. This is shown in the figure above.

Important cases

We ,here, consider few interesting cases :

Phase difference is zero

Zero phase difference means that individual SHMs are in phase with respect to each other. If we compare two SHMs as if they are independent and separate, then they reach mean position and respective “x” and “y” extremes at the same time. We can find the path of resulting motion by putting phase difference zero in the equation of path. Here,

x 2 A 2 + y 2 B 2 2 x y cos 0 0 A B = sin 2 0 0 x 2 A 2 + y 2 B 2 2 x y cos 0 0 A B = sin 2 0 0

x 2 A 2 + y 2 B 2 2 x y A B = 0 x 2 A 2 + y 2 B 2 2 x y A B = 0

x A y B 2 = 0 x A y B 2 = 0

y = B A x y = B A x

Figure 4: The path of motion is a straight line.
Lissajous curves
 Lissajous curves  (cs4.gif)

This is the equation of a straight line. The path of motion in reference to bounding rectangle is shown for this case. It is worth pointing here that we can actually derive this equation directly simply by putting φ = 0 in displacement equations in “x” and “y” directions and then solving for "y".

Clearly, motion of the particle under this condition is an oscillatory motion along a straight line. We need to know the resultant displacement equation. Let “z” denotes displacement along the path. By geometry, we have :

z = x 2 + y 2 z = x 2 + y 2

Substituting for “x” and “y” with φ = 0,

z = A 2 sin 2 ω t + B 2 sin 2 ω t = A 2 + B 2 sin ω t z = A 2 sin 2 ω t + B 2 sin 2 ω t = A 2 + B 2 sin ω t

This is bounded periodic harmonic sine function representing SHM of amplitude A 2 + B 2 A 2 + B 2 and angular frequency “ω” – same as that of either of the component SHMs.

Phase difference is π/2

A finite phase difference means that individual SHMs are not in phase with respect to each other. If we compare two SHMs as if they are independent and separate, then they reach mean position and respective “x” and “y” extremes at different times. When one is at the mean position, other SHM is at the extreme position and vice versa. We can find the path of resulting motion by putting phase difference “π/2” in the equation of path. Here,

x 2 A 2 + y 2 B 2 2 x y cos π 2 A B = sin 2 π 2 x 2 A 2 + y 2 B 2 2 x y cos π 2 A B = sin 2 π 2

x 2 A 2 + y 2 B 2 = 1 x 2 A 2 + y 2 B 2 = 1

Figure 5: The path of motion is an ellipse.
Lissajous curves
 Lissajous curves  (cs5.gif)

This is an equation of ellipse having major and minor axes “2A” and “2B” respectively. The resulting motion of the particle is along an ellipse. Hence, motion is periodic, but not oscillatory. If A = B, then the equation of path reduces to that of a circle of radius “A” or “B” :

Figure 6: The path of motion is a circle.
Lissajous curves
 Lissajous curves  (cs6.gif)

x 2 + y 2 = A 2 x 2 + y 2 = A 2

Phase difference is π

Individual SHMs are not in phase with respect to each other. When one is at one extreme, other SHM is at the other extreme position and vice versa. We can find the path of resulting motion by putting phase difference “π” in the equation of path. Here,

x 2 A 2 + y 2 B 2 2 x y cos π A B = sin 2 π x 2 A 2 + y 2 B 2 2 x y cos π A B = sin 2 π

x 2 A 2 + y 2 B 2 + 2 x y A B = 0 x 2 A 2 + y 2 B 2 + 2 x y A B = 0

x A + y B 2 = 0 x A + y B 2 = 0

y = B A x y = B A x

Figure 7: The path of motion is a straight line.
Lissajous curves
 Lissajous curves  (cs7.gif)

This is the equation of a straight line. The path of motion in reference to bounding rectangle is shown for this case. It is worth pointing here that we can actually derive this equation directly by putting "φ = π" in displacement equations in “x” and “y” directions.

Clearly, motion of the particle under this condition is an oscillatory motion along a straight line. We need to know the resultant displacement equation. Let “z” denotes displacement along the path. By geometry, we have :

z = x 2 + y 2 z = x 2 + y 2

x = A sin ω t x = A sin ω t

y = B sin ω t + φ = B sin ω t + φ = - B sin ω t y = B sin ω t + φ = B sin ω t + φ = - B sin ω t

Substituting for “x” and “y” with φ = π,

z = { A sin ω t 2 + - B sin ω t 2 } z = { A sin ω t 2 + - B sin ω t 2 }

z = A 2 sin 2 ω t + B 2 sin 2 ω t = A 2 + B 2 sin ω t z = A 2 sin 2 ω t + B 2 sin 2 ω t = A 2 + B 2 sin ω t

This is bounded periodic harmonic sine function representing SHM of amplitude A 2 + B 2 A 2 + B 2 and angular frequency “ω” – same as that of either of the component SHMs.

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