The Geometrical Interpretation Of Simple Harmonic Motion

Periodic motion is any motion that repeats itself within a regular interval, and the duration after which the motion repeats is called its time period. The journey of planets in the solar system and the action of a pump inside a cylinder are two examples of periodic motion around us. Simple harmonic motion (SHM) is when the displacement from the mean point is directly proportional to the quantity of the forces exerted on an oscillating body. We will see about the geometrical interpretation of SHM here after a basic understanding.

Basic Characteristics of SHM

Let’s specify some of SHM’s basic properties:

1. Amplitude (A): It is the measurement of the initial displacement.

2. The system that executes SHM switches between kinetic and potential forms of energy. The kinetic energy is zero at the extremes of the oscillations because the velocity is zero, and the potential energy is highest.

3. Time period (T): It is the time a particle takes for one oscillation. In most cases, the time period is measured either from the mean position or the extreme ends.

4. Frequency: It is the inverse of the time period and represents the number of entire oscillations executed every second., i.e.  v=1/T

SI unit: sec^-1 or hertz (Hz).

5. Angular frequency (ω), which is given by-

            ω  =2 πv

SI unit: radian per second

Differential Equation Of SHM

Let us assume a particle having mass (m) along the x-axis. Having its mean position at O, the particle is in simple harmonic motion.

Let’s assume that v_o = the speed at any point P.

At t = 0, the particle is going to the right from the mean point at point P.

At t = t, at point Q at a distance x.

We have the restoring force here as follows:

 F= -kx………………………….(1)

The final motion is simple harmonic because F is proportional to the displacement and opposite in direction to the displacement.

Here, k = the force constant. The SI unit of k is Nm^-1 .

By Newton’s second law of motion, expressing the equation (1) in the differential form:

F=ma=m d²(x)/dt² 

Therefore, the force here is:

      F=m d² (x)/dt² = -kx

On arranging, we can write it as follows:

       m d²(x)/dt² +kx=0

Or, d² (x)/dt² +k/m x=0

Or, d²(x)/dt² +ω²x=0…………………..(2)

The equation (2) above is the differential equation of SHM.

Here, ω = k/m is the angular frequency.

Solutions of Differential Equations of SHM

The differential equation has the solution:

1. x = A sin(ω t)

             ( particle at the mean point)

2. x_o = A sin Φ

( particle at a point other than the mean point)

3. x = A sin(ωt+Φ)

( particle at any position at time t)

These answers can be confirmed by inserting these solutions in equation(2).

Time Period of SHM

In solution (3), ω is observed as the coefficient of time (t).

We can write T of SHM for a particle as:

       T=2π/ω

or, ω=2π/T=2πf.

Here, angular frequency of particle = ωt,

and, f = frequency.

The velocity of a particle executing SHM

For a particle in SHM, the velocity (v) is written to be-

 v = dx/dt,

Since, we know that here x = A sin(ωt+Φ)

Therefore, v = d(x)/dt sin(ωt+Φ) = ω A cos(ωt+Φ),

              Or, v = Aω√1-sin²(ωt).

Since, x = A sin(ωt+Φ)

v = Aω√1-x²/A²,

v = ω√A²-x².

We will now square both sides-

v² = ω²(A²-x²),

Now, by dividing by ω² A² on both sides, we can obtain-

v² A²+v² / A² ω² =1.

It represents an equation of an ellipse.

The geometrical interpretation of SHM

When a particle moves with constant speed around a circle, there is a linear motion of the foot of the perpendicular of the moving particle, which is in SHM.

SHM as a projection of circular motion goes as follows:

Let us consider a particle revolving along a circle with angular velocity ω( ω is constant),

at t = 0, P is the position of the particle,

at t = t, Q is the position of the particle.

Here, the projection of P on the diameter = M

And the projection of Q on the diameter is = N

The projection M will move in a to-and-fro manner along the diameter as the particle P moves in an anti-clockwise direction.

Displacement of the projection = x-component of the vector(A).

x = A cos(ωt+Φ) …………….(1)

y = A sin(ωt+Φ) …………….(2)

So, P is in uniform circular motion.

Also, we see that (K and L) and (M and N) are in SHM about O.

So, Vector a_C = Aω² (towards the centre)

 a_C can be broken down into two components:

a_N = (ωt+Φ) Aω²sin² 

a_L = (ωt+Φ) Aω²cos²

a_N and a_L are the acceleration toward points N and L.

Here, horizontal phasor = foot of projection on the x-axis, and vertical phasor = the foot of projection on the y-axis.

Both the phases of the SHM have a difference of π/2.

Conclusion

We’ve looked at simple harmonic motion and its basic properties. In the case of SHM, we looked at the restoring force that occurs due to a deviation from the mean or equilibrium position and how it relates to the deviation’s amount. We constructed the equation of motion for SHM and calculated the result of the differential equation utilised to describe SHM using Newton’s second law of motion. SHM’s time period and angular frequency were also calculated.