Question

Explain Simple Harmonic Effect

Answer 1

In mechanics and physics, simple harmonic motion is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. ... The motion is sinusoidal in time and demonstrates a single resonant frequency.

Answer 2

Hi

Here is your answer,

A special type of periodic motion in which a particle moves to and fro     repeatedly about a mean position under the influence of a restoring force is know as Simple Harmonic Motion(SHM) or i can say Simple Harmonic Effect.

CHARACTERISTICS OF SIMPLE HARMONIC MOTION :-

1) DISPLACEMENT:- The displacement of a particle executing SHM at an instant is given by the distance of the particle from the mean position at that instant.

The values of displacement as a continuous function of time can be represented as a graph given below

Here,     Displacement, x (t) = A cos (ωt + Ф)

The function containing sine or cosine term is known as Sinusoidal function. 

2) AMPLITUDE:- The magnitude of maximum displacement of a particle executing SHM is called amplitude of the oscillation of that particle. Amplitude is measured on either side of mean position.

The displacement varies between the extremes +A and -A because the sinusoidal function of time varies from +1 to -1.

3) Phase :- If amplitude A is fixed for a given SHM then the state of motion position and velocity, of the particle at any time t is determined by the argument (ωt- Ф₀) in the sinusoidal function. The quality (ωt +Ф₀) is called phase of the motion.

For t = 0 phase ωt + Ф = Ф₀. Thus Ф₀ is called phase constant or phase angle . Two SHM may have the same 1 ( amplitude) and ω ( angular frequency) but different phase angle Ф.

As in the 3rd graph , the curve 1 has phase constant of (π/2  Ф = π/2)  and amplitude of A , the curve 2 has phase constant of π/4 ( Ф = +π/4 ) and amplitude of A.


4) Angular Frequency :- Angular frequency of a body executing motion is equal to the product of frequency of the particle with factor 2π. 

It is denoted by ω and it's SI unit is radian per second.

We know that,

                                 T = 2π/ω
                        
                                 ω = 2π/T = 2 π∨


where , ∨ = frequency of the particle.

Since, the motion is periodic with a period T, the displacement x (t) must return to it's initial value after one period of the motion x (t) must be equal to x ( t + T) for all t.

Consider the equation , x(t) = A cos ωt

                                         A cos ωt = A cos ω ( t+T ) 

Now, the cosine function is periodic with period 2π it first repeats itself when the argument changes by 2 π. Therefore,

                             → ω (t + T) = ωt + 2π

                             → Angular frequency , ω = 2π/T 

Thus , the angular frequency ω is 2π  times the frequency of oscillation 1/T .

Two SHM may have the same amplitude (A) and phase angle (Ф) but different angular frequency ω.

In the 4th diagram curve (b) has half the period and twice the frequency of the curve (a).


5) VELOCITY:- The velocity of a particle executing SHM at any instant, is defined as the time rate of change of it's displacement at that instant.

                  Velocity, ∨ = ω√A² - x²


At mean position , x = 0 
                                                          [ maximum velocity ]

                              v = ωA 

At extreme position   x = A 
                                                                   [ minimum velocity ]

                                     v = 0    


The velocity can also be calculated as 

                               v(t) = d/dt [ x (t) ] = d/dt [A cos (ωt + Ф)

                               Velocity, v (t) = -ωA sin (ωt + Ф)


6) Acceleration:- The acceleration of the particle executing SHM at any instant is defined as the time rate of change of its velocity at that instant.

                                Acceleration, a = -ω²x


At mean position, x = 0 
                                                                  [ minimum acceleration]
                                a = 0 


At extreme position  x= A
                                                                       [ maximum acceleration ] 
                                    a = -ω²A 

So, now the acceleration can also be calculated as 

                                    a (t) = d/dt [ v(t) ]

                                          = d/dt [ -ωA sin (ωt + Ф)


                  ∴ Therefore,  a (t) = -ω² A cos (ωt +Ф)




Hope it helps you !