Grade 11 Physics Note

Simple harmonic motion

Simple Harmonic Motion:

It is an oscillatory motion (to and fro) in which acceleration of a body is directly proportional to displacement and acceleration is always directed towards fixed point. (Mean position)

I.e. a $ \propto $ y where, ‘a’ is acceleration and ‘y’ is displacement.

Or, a = -k y, -ve sign show that ‘a’ is opposite to y

Displacement(y):

It is defined as the distance from mean position (XX1 of O) of a body executing SHM. It is denoted by ‘y’.

In right angle $\Delta $OPM, sin$\theta $ = OM/OP = y/r or, y = rsin$\theta $

We know, angular velocity, $\omega $ = $\frac{\theta }{{\rm{t}}}$∴$\theta $ =${\rm{\: }}\omega $t

Putting the value of $\theta $ in eqni) we get,

y = r sin${\rm{\: }}\omega $t

Velocity (v):                    

It is defined as the rate of change of displacement. It is denoted by v and given by $\theta $ = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$ -------i)

As we have, y = r sin$\omega $t ---------ii)

∴ From eqni) v = $\frac{{\rm{d}}}{{{\rm{dt}}}}$ (rsin$\omega $t)

= $\mu \frac{{\rm{d}}}{{{\rm{dt}}}}$ (rsin$\omega $t)

= $\mu \omega $cos$\omega $t

= $\omega \sqrt {{\mu ^2} - {\mu ^2}{\rm{sin}}} \omega $t

v = $\omega \sqrt {{\mu ^2} - {\mu ^2}} $

At mean poition, y = 0

Then,

v = $\omega \sqrt {{\mu ^2} - 0} $

v = $\omega \mu $

vmax = $\omega \mu $ at mean position

At extreme position, y =$\mu $

v = $\omega \sqrt {{\mu ^2} - {\mu ^2}} $

vmin = 0 at extreme position.

Acceleration (a):

It is defined as the rate of change of velocity. It is denoted by ‘a’ and is given by, a = $\frac{{{\rm{dv}}}}{{{\rm{dt}}}}$

As we know, v = $\frac{{{\rm{dy}}}}{{{\rm{dt}}}}$∴ a = $\frac{{\rm{d}}}{{{\rm{dt}}}}$$\left( {\frac{{{\rm{dy}}}}{{{\rm{dt}}}}} \right)$

As we know, y = y = rsin$\omega $t 

a = $\frac{{\rm{d}}}{{{\rm{dt}}}}$ [$\frac{{\rm{d}}}{{{\rm{dt}}}}$rsin$\omega $t ] = $\frac{{\rm{d}}}{{{\rm{dt}}}}$ [r$\frac{{\rm{d}}}{{{\rm{dt}}}}$ sin$\omega $t ]

a = $\frac{{\rm{d}}}{{{\rm{dt}}}}$ [r$\omega $cos$\omega $t ] ($\frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{\: }}$sinax = acosx)

a = r$\omega \frac{{\rm{d}}}{{{\rm{dt}}}}$ (cos$\omega $t)

a = r$\omega $(-$\omega $) sin$\omega $t     ($\frac{{\rm{d}}}{{{\rm{dx}}}}$cosax = -asinax)

a = $\omega $2rsin$\omega $t

a = $\omega $2y

At mean position, y = 0

Then, a = -$\omega $2${\rm{*}}$0

amin = 0

At extreme position, y = r

Then, a = -$\omega $2r

amax = $\omega $2r

Time period (T):

It is defined as the time required completing one oscillation. It is denoted by ‘T’. As we have, the magnitude of acceleration of a body executing SHM is,

a = $\omega $2y

a = ${\left( {\frac{{2{\rm{\pi }}}}{{\rm{T}}}} \right)^2}$y

a = $\frac{{4{{\rm{\pi }}^2}}}{{{{\rm{T}}^2}}}$ y

T = $\sqrt {\frac{{4{{\rm{\pi }}^2}}}{{\rm{a}}}} $ y

 

That total energy of a body executing simple harmonic motion is independent of the position of the body during its motion:

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. Simple harmonic motion can serve as a mathematical model of a variety of motions, such as the oscillation of a spring. In addition, other phenomena can be approximated by simple harmonic motion, including the motion of a simple pendulum as well as molecular vibration. Simple harmonic motion is typified by the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law. The motion is sinusoidal in time and demonstrates a single resonant frequency. In order for simple harmonic motion to take place, the net force of the object at the end of the pendulum must be proportional to the displacement.

 

Necessary and sufficient conditions for a motion to be simple harmonic:

It is an oscillatory motion (to and fro) in which acceleration of a body is directly proportional to displacement and acceleration is always directed towards fixed point (Mean position). If a particle moves to and fro about a mean position in a straight line such that the acceleration is directed towards the mean position and is directly proportional to the displacement from that position for a motion to be simple harmonic.

 

Characteristics of SHM are:

If a body is executing SHM, then

i) Its acceleration is directly proportional to displacement i.e. a $ \propto $ y

ii) Its acceleration is always directed towards mean position i.e. a = -k y

iii) It is a periodic motion

 

Test for simple harmonic motion;

Test for simple harmonic motion there should be;

i) Its acceleration is directly proportional to displacement i.e. a $ \propto $ y

ii) Its acceleration is always directed towards mean position i.e. a = -k y

iii) It must be a periodic motion

Simple pendulum:

When a heavy metallic bob is connected at one end of a light, inextensible thread and the thread passes through a frictionless rigid support so that the bob can oscillate in a fixed vertical plane. Then the device is called simple pendulum.

 

Drawbacks of simple pendulum:

1) Heavy point mass, light and inextensible thread can’t be achieved in practice.

2) Air resistance can’t be ignored.

3) The relation for time period holds true only for small oscillation.

4) The component mgsin$\theta $ is not perfectly directed towards mean position.

 

Relation between uniform circular motion and S.H.M:

Uniform circular motion describes the movement of an object traveling a circular path with constant speed. The one-dimensional projection of this motion can be described as simple harmonic motion.

 

Simple pendulum:

When a heavy metallic bob is connected at one end of a light, inextensible thread and the thread passes through a frictionless rigid support so that the bob can oscillate in a fixed vertical plane. Then the device is called simple pendulum.

Let us consider a body of mass ‘m’ is connected at one end of a thread of length ‘l’ which passes through a rigid support at point O.

Here,

The point O on the rigid support is called point of suspension and c.g. of the bob is called point of oscillation.

The distance between point of suspension and oscillation is called effective length.

At extreme position, let, mg is the wt. of the body and T be the tension produced in the string. The component mgcos$\theta $ is balanced by the tension (T).

The component mgsin$\theta $ provides restoring force to move the bob towards mean position.

∴ f = -mgsin$\theta $ , -ve sign is for restoring force.

or, ma = -mg sin$\theta $

or, a = -g sin$\theta $

for small angle, sin$\theta $$ \approx $$\theta $

a = -g$\theta $  --------------i)

By trigonometry,

$\theta $ = $\frac{{{\rm{arc\: length}}}}{{{\rm{radius}}}}$ = $\frac{{{\rm{AB}}}}{{{\rm{OA}}}}$

$\theta $ = $\frac{{\rm{y}}}{{\rm{l}}}$  ----------------ii)

Putting the value of $\theta {\rm{\: }}$in eqn ii) we get,

a = -g $\frac{{\rm{y}}}{{\rm{l}}}$

a = -($\frac{{\rm{g}}}{{\rm{l}}}$)y  ---------------iii)

Here, acceleration is directly proportional to displacement and they are opposite to each other.

Hence, motion of a simple pendulum is SHM

Expression for time period

If a body is in SHM then its acceleration is,

a = -${\omega ^2}$y -----------------iv)

Comparing eqn iii) and iv)

${\omega ^2}$ = $\frac{{\rm{g}}}{{\rm{l}}}$

or, $\omega $ = $\sqrt {\frac{{\rm{g}}}{{\rm{l}}}} $

or, $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ = $\sqrt {\frac{{\rm{g}}}{{\rm{l}}}} $  ($\omega $ = $\frac{{2{\rm{\pi }}}}{{\rm{T}}}$ )

or, T = 2${\rm{\pi }}\sqrt {\frac{{\rm{l}}}{{\rm{g}}}} $

 

A particle executing S.H.M., the maximum potential energy in same as that of maximum kinetic energy which is equal to total energy:

Energy in simple harmonic motion:

Kinetic energy;

                                                ${{\rm{E}}_{\rm{k}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^2}$

                          $ = \frac{1}{2}{\rm{m}}{{\rm{w}}^2}\left( {{{\rm{r}}^2} - {{\rm{y}}^2}} \right)$

Potential energy;

                                                ${{\rm{E}}_{\rm{p}}} = \frac{1}{2}{\rm{k}}{{\rm{y}}^2}$

                                                      $ = \frac{1}{2}{\rm{m}}{{\rm{w}}^2}{{\rm{r}}^2}$

Total energy of the particle at any point is

E = ${{\rm{E}}_{\rm{p}}} + {{\rm{K}}_{\rm{E}}}$

                           $ = 2{\rm{\: m\: }}{{\rm{r}}^2}{{\rm{f}}^2}{\rm{\: \: }}{{\rm{r}}^2}{\rm{\: }}$

i. When particle is at mean position, y = o

Then energy is equal to the maximum value of K.E.

ii. When particle is at extreme position, y = r

And total energy is equal to the maximum value of P.E.
 

Mass spring system and time period for the mass spring system:

As we know,

                                                                     ${\rm{a}} =  - {\omega ^2}{\rm{l}}$

This shows the acceleration is directly proportional to the displacement and is directed towards the mean position. Hence, the motional of a horizontal mass spring system is simple harmonic motion.

 We can derive the time period for the mass spring system by as we know

                                                             ${\omega ^2} = \frac{{\rm{k}}}{{\rm{m}}}$

Where $\omega $ is the angular velocity, if T is the time of oscillation, then

                                                    $\omega {\rm{\: }} = \frac{{2{\rm{\pi }}}}{{\rm{T}}}$

$\omega {\rm{\: }} = \frac{{2{\rm{\pi }}}}{{\rm{T}}}$ = $\sqrt {\frac{{\rm{K}}}{{\rm{m}}}} $

Here, T= $2{\rm{\pi }}\sqrt {\frac{{\rm{m}}}{{\rm{K}}}} {\rm{\: }}$

This is the required expression for the time period of a mass-spring system. The mass of the spring is usually neglected in such type of problem because this expression shows that the time period of oscillation depends upon the attached to the spring.


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