Simple Harmonic Motion And Oscillation

Simple Harmonic Motion And Oscillation

Simple Harmonic motion (SHM):

Any motion which repeats itself after regular interval of time is called periodic or harmonic motion.

If a particle in periodic motion moves back and forth (or to and fro) over the same path, then its motion is called oscillatory or vibratory. The examples of oscillatory or vibratory motion are:

1. the motion of a pendulum

2. the motion of a spring fixed at one end, which is stretched or compressed and then released

3. the motion of a violin string

4. the motion of atoms in molecules or in a solid lattice

5. the motion of air molecules as a sound wave passes by

Conditions of Simple Harmonic Motion

For SHM is to occur, three conditions must be satisfied.

1. There must be a position of stable equilibrium

At the stable equilibrium potential energy is minimum.

That is, $\frac{dU}{dx}=0$ and $\frac{d^{2}U}{d{x}^2}>0$

2. There must be no dissipation of energy

3. The acceleration is proportional to the displacement and opposite in direction.

That is, $a=-{\omega }^{2}x$

TYPES OF MOTION

(i) PERIODIC MOTION

When a body or a moving particle repeats its motion along a definite path after regular intervals of time, its motion is said to be Periodic Motion and interval of time is called time or harmonic motion period (T). The path of periodic motion may be linear, circular, elliptical or any other curve.

(ii) OSCILLATORY MOTION

'To and Fro' type of motion is called an Oscillatory Motion. It need not be periodic and need not have fixed extreme positions.

The force/torque acting in oscillatory motion (directed towards equilibrium point) is called restoring force/torque.

(iii) EQUATION OF SIMPLE HARMONIC MOTION (SHM):

The necessary and sufficient for a SHM = Force constant

F = –kx

where k = Force constant for a SHM,

x = displacement from mean position.

or $m\frac{d^{2}x}{d{t}^2}=-kx$

$\Rightarrow$ $\frac{d^{2}x}{d{t}^2}+\frac{k}{m}x=0$

$\Rightarrow$ $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0where\omega =\sqrt{\frac{k}{m}}$

It's solution is $x=A\sin \left(\omega t+\varphi \right)$

CHARACTERISTICS OF SHM

(i) Amplitude: It is the maximum value of displacement of the particle from its equilibrium position.

(ii) Time period (T): Smallest time interval after which the oscillatory motion gets repeated is called Time period.

(iii) Frequency (f): Number of oscillations competed in unit time interval is called frequency of oscillations, $f=\frac{1}{T}=\frac{\omega }{2\pi }$ its units is ${\sec }^{-1}$ or Hz.

(iv) Angular Frequency (w): The quantity $\omega$$\omega$ is called the angular frequency of the oscillating system. As we know that second order differential equation of simple harmonic motion $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0where\omega =\sqrt{\frac{k}{m}}$;is called angular frequency and its units is rad/sec.
(v) ;;;;;Phase: The physical quantity which represents the state of motion of particle (eg. its position and direction of motion (orientation);at any instant).In the solution of sedond order differential equation of SHM, $\varphi$ is called phase of the motion.(vi) ;;;;Displacement (x) ;;;;;;;;;;;If time is measured from the equilibrium;position ,;displacement x from equilibrium point at any instant of time t is given by x = Asin$\omega$t(vii) ;;;Velocity (v): v = $\frac{dx}{dt}=\frac{d}{dt}$(A sin$\omega$t) = $\omega$A cos$\omega$t;;;;;;;;;;;;;;;;;;;;;;;;;;;;;or;;v = $\omega$A$\sqrt{1-{\sin }^2\omega t}$ = $\omega$$\sqrt{A^2-x^2}$

(a) Velocity is minimum at extreme positions and is zero.

At x = A, v = vmin = zero.

(b) Velocity is maximum at the equilibrium position and is $\omega$A.

At x = 0, v = vmax = $\omega$A

(c) Direction of velocity is either towards or away from the equilibrium position.

(viii) Acceleration (a) a = $\frac{dv}{dt}=\frac{d^{2}x}{d{t}^2}$ = -$\omega$2A sin$\omega$t = -$\omega$2x

(a) The minimum value of acceleration is zero and it occurs at equilibrium.

(b) The maximum value of acceleration is $\omega$2A and it occurs at extreme positions.

(c) Acceleration is always directed towards the equilibrium position and so it is always opposite to the direction of the displacement.

ENERGY OF A BODY IN S.H.M.

(i) Potential Energy

The linear restoring force acting on the harmonic oscillator is given by

F = m$\frac{d^{2}x}{d{t}^2}$ = – kx

Now if the oscillator is displaced through a further displacement dx against the force, work done in displacing the particle is given by

dW = kx dx

Hence the total work done in displacing the particle from mean position (x=0) to (x=x) is given by

W = ${\int }_0^{x}kxdx=\frac{1}{2}k{x}^2$

By convention P.E. at the mean position is taken as zero. Hence, above equation gives the values of P.E. of harmonic oscillator at a displacement x from the mean position i.e.,

U = $\frac{1}{2}k{x}^2=\frac{1}{2}m{\omega }^2{x}^2$ . . . (1)

This shows the P.E. is proportional to the square of the displacement and graph showing the variation of potential energy with the displacement will be a parabola given by continuous line in the figure. P.E. is maximum at maximum displacement and is given by

Umax = $\frac{1}{2}m{\omega }^2{A}^2$

(ii) Kinetic Energy

Velocity of harmonic oscillator is given by equation as

v = $\frac{dx}{dt}$ = $\omega \sqrt{A^2-x^2}$

Hence kinetic energy of the oscillator is given by

K.E. = $\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)$ . . . (2)

The graph showing the variation of K.E. with x is shown in figure by dotted line.

The kinetic energy is maximum when x = 0. Thus

K.E.max = $\frac{1}{2}m{\omega }^2{A}^2$

Now total energy E of the oscillator for displacement x is given by

E = P.E. + K.E.=$\frac{1}{2}m{\omega }^2{x}^2+\frac{1}{2}m{\omega }^2\left(A^2-x^2\right)$

= $\frac{1}{2}m{\omega }^2{A}^2$= constant . . . (3)

Thus total energy is independent of the displacement, it remains constant throughout the motion of the oscillator. Also the total energy is equal to maximum value of either P.E. or K.E.

(iii) Average Value of P.E. and K.E.

By equation (1) P.E. at distance x is given by

U = $\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\left(\omega t+\phi \right)$

The average value of P.E. for one complete oscillation is given by

$U_{average}=\frac{1}{T}\underset{0}{\overset{T}{\int }}Udt=\frac{1}{T}\underset{0}{\overset{T}{\int }}\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\left(\omega t+\phi \right)$dt

= $\frac{m{\omega }^2{A}^2}{4T}\underset{0}{\overset{T}{\int }}2{\sin }^2\left(\omega t+\phi \right)dt$

m=$\frac{1}{4}m{\omega }^2{A}^2$

Because the average value of sine or of cosine function for the complete cycle is equal to zero.

Now K.E. at x is given by

$T=\frac{1}{2}m{\left(\frac{dx}{dt}\right)}^2=\frac{1}{2}m{\left[\frac{d}{dt}\left\{A\sin \left(\omega t+\phi \right)\right\}\right]}^2$

= $\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\left(\omega t+\phi \right)$

The average value of K.E. for one complete cycle

Taverage = $\frac{1}{T}\underset{0}{\overset{T}{\int }}\frac{1}{2}m{\omega }^2{A}^2{\cos }^2\left(\omega t+\phi \right)dt$

= $\frac{m{\omega }^2{A}^2}{4T}\underset{0}{\overset{T}{\int }}\left\{1+\cos 2\left(\omega t+\phi \right)\right\}dt$

=$\frac{m{\omega }^2{A}^2}{4T}.T=\frac{1}{4}m{\omega }^2{A}^2$

Thus average values of K.E. and P.E. of harmonic oscillator are equal and each is equal to half of the total energy.

Exmaple : A particle executes S.H.M. with time period 4s find the time taken by the particle to go directly from its mean position to half its amplitude.

Solution: $x=A\sin \left(\omega t+{\varphi }_0\right)$

At $t=0,x=0\Rightarrow A\sin {\varphi }_0=0or{\varphi }_0=0$

Hence, $x=A\sin \left(\omega t\right)$

or $A/2=A\sin \left(\omega t\right)$

or $1/2=\sin \left(\omega t\right)$

$\omega t={\sin }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$

$t=\frac{\pi }{6\omega }=\frac{\pi T}{6\left(2\pi \right)}$

as $\omega =2\pi /T$

$\Rightarrow t=T/12=1/3s$