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## Superposition of Harmonic Oscillations, Physics tutorial

:Superposition of Simple Harmonic OscillationsWhen we superpose initial conditions corresponding to velocities and amplitudes, resultant displacement of two (or more) harmonic displacements will be algebraic sum of individual displacements at all subsequent times. The principle of superposition holds for any number of simple harmonic oscillations. These may be in same or mutually perpendicular directions, i.e. in two dimensions.

Given equation defines SHM:

d

^{2}x/dt^{2}= -ω_{0}^{2}x........................................Eq.1This is the linear homogeneous equation of second order. Such equation has significant property that sum of its two linearly independent solutions is itself a solution.

Let x

_{1}(t) and x_{2}(t) respectively satisfy equationsd

^{2}x_{1}/dt^{2}= -ω_{0}^{2}x_{1}........................................Eq.2d

^{2}x_{2}/dt^{2}= -ω_{0}^{2}x_{2}........................................Eq.3Then by adding Eq.2 and Eq.3, we get

d

^{2}(x_{1}+ x_{2})/dt^{2}= -ω_{0}^{2}(x_{1}+ x_{2}) ........................................Eq.4According to principle of superposition, sum of two displacements provided by

x(t) = x

_{1}(t) + x_{2}(t) ........................................Eq.5This also satisfies Eq.1. In other words, superposition of two displacements satisfies same linear homogeneous differential equation that is satisfied individually by x

_{1}and x_{2}.:Superposition of two harmonic oscillations of the same frequency along the same lineLet us superpose two collinear (along same line) harmonic oscillations of amplitudes a

_{1}and a_{2}having frequency ω_{0}and phase difference of π. Displacements of the oscillations are provided byx

_{1}= a_{1}cosω_{0}t.......................Eq.6And x

_{2}= a_{2}cos(ω_{0}t + π)= -a

_{2}cosω_{0}t.......................Eq.7According to principle of superposition, resultant displacement is provided by

x(t) = x

_{1}(t) + x_{2}(t)= a

_{1}cosω_{0}t - a_{2}cosω_{0}t= (a

_{1}- a_{2})cosω_{0}t.......................Eq.8This represents the simple harmonic motion of amplitude (a

_{1}- a_{2}). In particular, if two amplitudes are equal, i.e. a_{1}= a_{2}, resultant displacement will be zero at all times.Superposition of two collinear harmonic oscillations of different frequencies

In numerous cases, we have to deal with superposition of two or more harmonic oscillations having different angular frequencies. The microphone diaphragm and human eardrums are simultaneously subjected to different vibrations. For ease, we shall first consider superposition of two harmonic oscillations having same amplitude a but somewhat different frequencies ω

_{1}and ω_{2}such that ω_{1 }> ω_{2}:x

_{1}= acos(ω_{1}t + Φ_{1})x

_{2 }= acos(ω_{2}t + Φ_{2})Phase difference between the two harmonic vibrations is:

Φ = (ω

_{1}- ω_{2})t + (Φ_{1}- Φ_{2})The first term (ω

_{1}- ω_{2})t changes continuously with time. But second term (Φ_{1}- Φ_{2}) is constant in time and as such it doesn't play any important role here. Thus, we may assume that initial phase of two oscillations are zero. Then, two harmonic oscillations can be written as:x

_{1}(t) = acosω_{1}tx

_{2}(t) = acosω_{2}t................Eq.9The superposition of two oscillations gives the resultant

x(t) = x

_{1}(t) + x_{2}(t) = acos(ω_{1}t + ω_{2}t)This equation can be rewritten as

x(t) = 2acos((ω

_{1}- ω_{2})/2)tcos((ω_{1}+ ω_{2})/2)tThis is the oscillatory motion with angular frequency

(ω

_{1 }+ ω_{2})/2 and amplitude 2acos((ω_{1}- ω_{2})/2)tThe average angular frequency

ω

_{av}= (ω_{1 }+ ω_{2})/2 and the modulated angular frequencyωmod = (ω

_{1}- ω_{2})/2We find that amplitude a

_{mod}(t) = 2acosωmodt differs with the frequency ω_{mod}/2π = (ω_{1}- ω_{2})/4πThis also signifies that in one complete cycle modulated amplitude takes values of 2a, 0, -2a, 0 and 2a for ω

_{mod}t = 0, π/2, π, 3π/2 and 2π, respectively. Resultant oscillation can be written asx(t) = a

_{mod}(t)cosω_{av}tThis equation looks like the equation of SHM. But this resemblance is misleading. Modulated amplitude and phase constant are respectively provided by:

a

_{mod}(t) = [a_{1}^{2}+ a_{2}^{2}+ 2a_{1}a_{2}cos(2ωmodt)]^{1/2}And θ

_{mod}= [(a_{1}- a_{2})sinω_{mod}t/(a_{1}+ a_{2}cosω_{mod}t)]:Superposition of many harmonic oscillations of same frequency Method of Vector AdditionThis method is based on fact that displacement of the harmonic oscillation is projection of the uniform circular motion on diameter of circle. Thus, it is significant to understand connection between SHM and uniform circular motion.

Uniform Circular Motion and SHM:Let us assume that a particle moves in the circle with constant speed V. Radius vector joining centre of circle and position of particle on the circumference will rotate with the constant angular frequency. We take x -axis to be along the direction of radius vector at time t = 0. Then angle made by the radius vector with x -axis at any time t will be given by

θ = length of arc/radius of the circle = Vt/R

The x - and y - components of position of particle at time t are

x = Rcosθ and y = Rsinθ

Therefore, dx/dt = -Rsinθdθ/dt

= -ω

_{0}RsinθAs dθ/dt = ω

_{0}= V/RSimilarly, you can write

dy/dt = ω

_{0}RcosθDifferentiating again with respect to time, we get

d

_{2}x/dt_{2}= -ω_{0}^{2}x and d^{2}y/dt^{2}= -ω_{0}2yThese expressions show that when the particle moves uniformly in the circle, its projections along x - and y - axes execute SHM. A simple harmonic motion may be viewed as the projection of uniformly rotating vector on reference axis.

Assume that vector OP' with |OP'| = a

_{0}is rotating with angular frequency ω_{0}in anticlockwise direction, as shown in Figure given below. Let P be foot of perpendicular drawn from P' on x - axis. Then OP = x is projection of OP' on x -axis. As vector OP' rotates at constant speed, point P executes simple harmonic motion along x -axis. Its period of oscillation is equivalent to period of rotating vector OP'. Let OP_{0}' be initial position of the rotating vector. Its projection OP_{0}on x -axis is a_{0}cosΦ. If rotating vector moves from OP_{0}' to OP' in time t, then ∠P'OP_{0}' ω_{0}t and ∠P'_{O}= (ω_{0}t + Φ) Then we can writeOP = OP'cos∠P'

_{O}xor x = a

_{0}cos(ω_{0}t + Φ)Therefore, point P executes simple harmonic motion along x-axis. If you project OP' on y -axis, you will find that point corresponding to the foot of the normal satisfies equation

y = a

_{0}sin(ω_{0}t + Φ)This signifies that rotating vector can, in general, be resolved in two orthogonal components, and we can write

r = x

_{x}+ y_{y}Where x

_{x}and y_{y}are unit vectors along x - and y -axes, respectively.Vector Addition:Let us now consider superposition of n harmonic oscillations, all having same amplitude a

_{0}and angular frequency ω_{0}. Initial phases of successive oscillations differ by Φ_{0}. Let first of these oscillations be defined by equationx

_{1}(t) = a_{0}cosω_{0}tThen, other oscillations are provided by:

x

_{2}(t) = a_{0}cos(ω_{0}t + Φ_{0}).....x_{n}(t) = a_{0}cos[ω_{0}t + (n-1)Φ_{0}]From principle of superposition, resultant oscillation is expressed by

x(t) = a

_{0}[cosω_{0}t + cos(ω_{0}t + Φ) + cos(ω_{0}t + 2Φ_{0}) +......cos(ω_{0}t + (n-1)Φ_{0})]:Oscillations in two dimensionsOscillatory motion in two dimensions is also possible. Most recognizable example is motion of the simple pendulum whose bob is free to swing in any direction in x-y plane. This arrangement is known as spherical pendulum. We displace pendulum in x-direction and as we release it, we provide it impulse in the y-direction. The result is a composite motion when such a pendulum oscillates whose maximum x-displacement happens when y-displacement is zero and y velocity is maximum and vice versa. Remember that as the time period of the pendulum depends only on acceleration because of gravity and length of the cord, frequency of the superposed SHM's will be the same. Result is the curved path, in general, an ellipse. We now apply the principle of superposition to the case where two harmonic oscillations are perpendicular to each other.

:Superposition of Two Mutually Perpendicular Harmonic Oscillations of Same FrequencyLet two mutually perpendicular oscillations having amplitudes a

_{1}and a_{2}, such that a_{1}> a_{2}and angular frequency ω_{0}. These are explained by equationsx

_{1}= a_{1}cosω_{0}t and x_{2}= a_{2}cos(ω_{0}t + Φ)Here we have taken initial stage of vibrations along x and the y-axes to be zero and Φ respectively. That is, Φ is a phase difference between two vibrations. First determine resultant oscillation for few particular values of phase difference

Case 1: Φ = 0 or π

For Φ = 0 x = a

_{1}cosω_{0}t and y = a_{2}cosω_{0}tTherefore y/x = a

_{2}/a_{1}or y = (a_{2}/a_{1})xLikewise for Φ = π x = a

_{1}cosω_{0}t and y = -a_{2}cosω_{0}tSo that y = -(a

_{2}/a_{1})xThe equation of y given above explains straight lines passing through origin. This signifies that resultant motion of particle is along straight line. Though for Φ = 0 the motion is along one diagonal but when Φ = π the motion is along the other diagonal.

Case 2: Φ = π/2. In this case two vibrations are provided by

x = a1cosω0t

y = a

_{2}cos(ω_{0}t + π/2) = -a_{2}sinω_{0}tOn squaring the expressions and adding resultant expressions, we get

x

^{2}/a_{1}^{2}+ y^{2}/a_{2}^{2}= cos2Φ + sin^{2}Φ = 1This is equation of ellipse. Therefore resultant motion of particle is along ellipse whose principal axes lie along x- and y-axes. Semi-major and semi-minor axes of ellipse are a

_{1}and a_{2}. As time increases x decreases from maximum positive value but y becomes more and more negative. Therefore, ellipse is explained in clockwise direction. If you analyze case when Φ = 3π/2 or Φ = π/2, you will get same ellipse, but motion will be in anticlockwise direction.When amplitudes a

_{1}and a_{2}are equal, a_{1}= a_{2}= a Equation given above reduces tox

^{2}+ y^{2}= a^{2}This equation represents the circle of radius a. This signifies that ellipse reduces to circle. This equation represents the circle of radius a. This denotes that ellipse reduces to the circle.

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