simple harmonic motion

We know that rectilinear motion and motion of projectiles are non-repetitive, but planetary motion in the solar system and uniform circular motion are repetitive. Simple harmonic motion deals with the repetitive nature of the motion of a particle, i.e.”a motion in which restoring force is directly proportional to displacement of the body by its mean position”.

PERIODIC AND OSCILLATORY MOTION

• A motion that repeats itself at regular intervals of time is called periodic motion.

  Example:- When you play the game of bouncing the ball off the ground, between your palm and ground.

• If the body is given small displacement from the position, a force comes into play which tries to bring the body back  to the equilibrium point, giving rise to 

oscillations or vibrations.

Example:-A ball placed in a bowl will be in equilibrium at the bottom. If displaced little from the point, it will perform oscillations in the bowl.

• Every oscillatory motion is periodic motion, but converse is not true.

Example:- circular motion is periodic but not oscillatory.

PERIOD AND FREQUENCY

The smallest interval of time after which the motion is repeated is called its period. Its SI unit is second.

If T is period then the reciprocal of T i.e is 1/T is called frequency of periodic motion.

Its SI unit is sec-1.

FREQUENCY AND ANGULAR FREQUENCY

The reciprocal of period is called the frequency, or one can also say that frequency represents the no. of oscillation per unit time. It is measured in hertz, i.e. cycles per second, and denoted as Hz.

Mathematically v=1÷T=2T constant is called the angular frequency.

SIMPLE HARMONIC MOTION

Consider a particle of mass m moving along the X-axis. Suppose a force F=-kx acts on the particle where k is a positive constant and x is the displacement of the particle from the assumed origin. The particle then executes a SHM with the centre of oscillation at the origin.

Let the position of the particle at t=0 is x0 and its velocity is v0.

Then at t=0,  x= x0   and    v=v0

And acceleration of the particle at any instant is 

a=FM=-KxM=-2x

  Where,  =kM

Thus dvdt=–2x

Ordvdx dxdt=–2x

Or v(dvdx) =–2x

Or vdv=–2xdx

when the particle is at x=x0 velocity is v0

Integrating above equation taking limit from v0to v for velocity and

distance x0 to x we get,

v2–v02=–2(x2–x02)

Or  v2=v02–2x2+2x02

or v=v02–2x2+x02

Or  v=v022–x2+x02  

Writing v022+x02=A2

Now above equation can be written as 

   v=A2–x2

Above equation can be expressed as 

   dxdt=A2–x2

Or  dxA2–x2=dt

 when t=0 ,the displacement is x0 and at time t the displacement becomes x.

Integrating above equation x=x0 to x and t=0 to t

  sin-1(xA)-sin-1(x0A)=t

Or sin-1(xA)=sin-1(x0A)+t

Or xA=sin{t+sin-1(x0A)}

Or x=Asin{t+sin-1(x0A)}

 Or x=Asin(t+)………………………..(1)

Where =sin-1(x0A) is called phase difference 

The velocity at time t is 

v=Acos(t+)……………………..(2)

AMPLITUDE

Equation 1 gives the displacement of a particle in shm. As  sin(t+) is between -1 to 1, the displacement takes -A to A.

TIME PERIOD

 We have x=Asin(t+)

If T be the time period, then x should have same value at t and t+ T

So sin(t+)=sin[(t+T)+]

Now velocity is v=Acos(t+)

As the velocity also repeat it’s value after a time period 

cos(t+)=cos[(t+T)+]

Both sin(t+) and cos(t+) will repeat their value if the angle (t+)increases by 2 or its multiple.

So [(t+T)+] =(t++2)

 Or  T=2

We know =kM

 So,   T=2

Or    T=2Mk…..……………(3)

Where k is the force constant and M is the mass of the particle.

Example:-A particle of mass 400 g executes a shm.The restoring force is provided by a spring of spring constant 160N/m.Find the time period.

Sol:-Given mass M= 400g=0.4kg

                  Spring constant K= 160N/m

 The formula for time period can be traced back to equation 3 using equation 3 we can write 

 T=20.4160

     T= 2π × 0.05=0.31s

MATHEMATICAL EXAMPLE

1:The equation of a particle executing simple harmonic motion is 

x=5sin(t+3) .Write down the amplitude,time period and maximum speed.Also find velocity at t=1s.

Sol:- we know that x=Asin(t+),

So comparing with that in the question 

we will get amplitude=5m

And time period = 2π÷

                          =2π÷π

                          =2sec

The maximum speed = A×

                                  =5 ×π

                                 = 5 m/s

Velocity can be expressed v=Acos(t+)

   At t=1s

        v=(5)(π)cos(53)

          =-52

3:what will be the state of force, acceleration, velocity, kinetic energy and potential

energy when a particle executing SHM passes through the mean position?

Ans: No force will act on the particle, acceleration of the particle is zero, velocity is maximum, kinetic energy is maximum, potential energy is zero.

4:Question is the same as previous, with the only difference being when SHM is at the end?

Ans: Acceleration of particle is maximum,restoring force acting on the particle is maximum,velocity is zero,kinetic energy of particle is zero,potential energy is maximum.

CONCLUSION:-

In science and engineering there are many uses of SHM. This study material contains definition of SHM, and from the derivation we conclude the formulas: 

The equation of simple harmonic is given, x=Asin(t+) .

The velocity at time t is v=Acos(t+). 

The equation of time period is  T=2

Or    T=2Mk