Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What is a Sinusoidal Function?
If a particle executing simple harmonic motion begins its motion from the position of equilibrium, then the equation of motion of the particle is
x = Asinωt = Asin\(\frac{2 \pi}{T}\)t
During a complete period, i.e., from t = 0 to t = T, the variation of displacement with time is shown in the following table:
Now on a graph paper, time is placed along the horizontal axis and displacement along the vertical axis. The points O, B, C D and E are plotted according to their corresponding coordinates in the above table. The points are connected with a curved line. The graph OBCDE is obtained [Fig.]. This is a sine curve. This curve represents a simple harmonic motion.
If a particle executing simple harmonic motion starts its motion from one end of its path then the equation of motion of the particle is
x = Acosωt = Acos\(\frac{2 \pi}{T}\)t
Proceeding similarly, we get the following table and the graph [Fig.]. The graph B’C’D’E’F’ is a cosine curve.
The sine and the cosine curves are identical except for their initial phase difference. For their special symmetry sine and cosine functions are called sinusoidal functions. As a simple harmonic motion is represented by any one of these sinusoidal functions, they are also known as harmonic functions.
Composition of Two Collinear SHMs of the Same Frequency by Graphical and Analytical Methods
Two simple harmonic motions, having the same frequency n (i.e., time period, T = \(\frac{1}{n}\)) and angular frequency ω = 2πn, are being executed along the x -axis, i.e., they are collinear. To obtain the resultant motion due to superposition of the two SHMs, we have to know their phase difference.
1. Two SHMs are in phase: The two SHMs of the same time period are in phase.
Let the equation of the first motion be
x1 = Asinωt = Asin\(\frac{2 \pi}{T}\)t.
and the equation of the second motion be
x2 = Bsinωt = Bsin\(\frac{2 \pi}{T}\)t.
The resultant displacement at any instant is the vector sum of the displacements of the two individual motions at that
instant.
So, the resultant displacement.
x = x1 + x2 = A sin\(\frac{2 \pi}{T}\) + B sin\(\frac{2 \pi}{T}\) = (A + B) sin\(\frac{2 \pi}{T}\)
Therefore, the resultant motion is an SHM of the same frequency with an amplitude equal to (A + B).
The displacements of the two individual motions and their resultant displacement at different times (from t = 0 to
t = T) are shown in the table below.
Plotting the referred values in the table, we get the graph of the resultant displacement x [Fig.]. From the graph we come to know
- The resultant motion is also simple harmonic.
- The frequency and the time period of the resultant motion are equal to those of the individual motions.
- The amplitude of the resultant motion is equal to the sum of the amplitudes of the individual motions.
- The resultant motion is also in phase with the individual motions.
2. Two SHMs are in a opposite phase: Two simple harmonic motions have the same frequency and the same time period, but they are in opposite phase.
So, the equation of the first SHM:
x1 = Asinωt = Asin\(\frac{2 \pi}{T}\)t.
The equation of the second SHM:
x2 = Bsin(ωt + π) = -Bsinωt = -Bsin\(\frac{2 \pi}{T}\)t.
[As the two motions are in opposite phases, phase difference between them = 180° = π]
∴ Resultant displacement,
x = x1 + x2 = Asin\(\frac{2 \pi}{T}\)t – Bsin\(\frac{2 \pi}{T}\)t = (A – B)sin\(\frac{2 \pi}{T}\)t
From t = 0 to t = T, the values of x1, x2 and x are given in the table below:
Plotting the referred values in the table, we get the graph of the resultant displacement x [Fig.]. From the graph we come to know
- The resultant motion is also simple harmonic.
- The frequency and the time period of the resultant motion are equal to those of the individual motions.
- The amplitude of the resultant motion is equal to the difference of the amplitudes of the individual motions.
- The resultant motion is in phase with the motion having larger amplitude, but in opposite phase with the motion having smaller amplitude.
Due to the superposition of two SHMs of the same amplitude but of opposite phase, the resultant displacement is x = x1 + x2 = 0 for all values of t. It implies that the particle, on which these two SHMS superpose, remains at rest.