A system executing simple harmonic motion is called a simple harmonic oscillator. In simple harmonic motion, the force acting on the system at any instant, is directly proportional to the displacement from a fixed point in its path and the direction of this force is towards that fixed point. Thus, the system executes the motion under a linear restoring force. If the displacement of the system from a fixed point is x, the linear restoring force is -Kx, where K is a constant which is called the force constant. Thus no other force except the linear restoring force acts on a simple harmonic oscillator. As a result, the oscillator executes vibrations of constant amplitude and with a constant frequency. These oscillations are called the free oscillations.

Let a particle of mass m be executing simple harmonic oscillations. The acceleration of the particle at displacement x from a fixed point will be d^{2}x/ dt^{2} . For the particle, restoring force α-displacement.

where K is a constant, which is called force constant of the particle. Here the negative sign tells that the direction of force acting on the particle (or acceleration) is opposite to the direction of increase in displacement (Figure 1.2).

FIGURE 1.2 Direction of restoring force

Acce1eration of the particle d^{2}x /dt^{2} = Kx /m

Let k/m =w^{2}

then

Acceleration of the particle d^{2 x }/dt^{2} = – w^{2}x

Equation (1) is known as differential equation of simple harmonic oscillator.

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**1.3.1 Solution of Differential Equation of Simple Harmonic Oscillator**

Now we have to find the displacement x of the particle at any instant t by solving the differential equation (1) of the simple harmonic oscillator.

In equation (1), multiplying by 2( dx/dt),we get

At the position of maximum displacement, i. e., at x =±a, ve1 o City of particle dx/dt = 0

0 + w^{2}a^{2} =A or A =-w^{2}a^{2}

Then (dx^{2} /dt^{2})+ w^{2} x^{2 } = w^{2 }a^{2}

Hence ( dx/dt)^{2}=w2 (a^{2}-x^{2})

or velocity of the particle dx/ dt =w /(a^{2}-x^{2})

Equation (2) tells us the velocity of particle at position x

From equation (2) dx /(a^{2}-x^{2}) –w dt

Again integrating, sin-^{1 }x/a = wt + φ(contant)

x/a =sin ( wt+ φ) or x=a sin (wt + φ)

Here a is the amplitude of oscillations and φ is the initial phase of the motion of particle (whose value can be known from the initial conditions).

(i) If x = 0 at t = 0 (i.e. the particle_ initiates its oscillations from its mean position), then sin φ = 0 or φ = 0, then the displacement equation of the particle executing simple harmonic motion at any time t will be

x =a sin *ω*t

(ii) If x = 0 at t = 0 (i.e. , the particle initiates its oscillations from its maximum displaced position), then sin φ = 1 or φ = *ω* / 2, then the displacement equation of the particle executing simple harmonic motion at any instant t will be x =a sin (*ω*t+ *ω* / 2) =a cos *ω*t

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**1.3.2 Time Period and Frequency**

It is clear from equations (4) and (5), that

a sin *ω* ( t + 2π/*ω*) = a sin *ω*t and a cos *ω* ( t + 2 π/*ω*) = a cos *ω*t

It is concluded from here that the displacement of the particle at any instant ( t + 2π/*ω*) is exactly the same as it was at the instant t i.e., the particle comes back to its initial position during its motion exactly after time it . Hence time period of the particle executing simple harmonic motion is

It is clear from equation (1) that numerically, acceleration = of x displacement

ω =

and time period T =2π

Since frequency *n *= 1/T, hence frequency *n *= 1/2π

Remember that the quantity *w *= *2nn *is known as the angular frequency of motion .