Waves and oscillation tebranishlar va to’lqinlar

A restoring force which is proportional to displacement but oppositely directed. This is written as -Kx, where µ is a constant of proportionality or force constant.

(ii) A resistive force proportional to velocity but oppositely directed. This may be written as

- r , where r is the frictional force per unit velocity.


^{RESISTIVE FORCE ( F}R^{) = −𝒓 𝒅𝒙 (𝟏)}
𝒅𝒕
^{Restoring force (F}x^{) = -Kx (2)}

By using newton’s 2^nd law
𝑭 = 𝒎 ^𝒅𝒙 … … … … … … … (𝟑)
𝒅𝒕
Now total force acting on a body is given by
F = RESISTIVE FORCE + RESTORING FORCE
Using equation (1), (2) and (3)

𝒅^𝟐𝒙
𝒅𝒙

𝒎 _𝒅𝒕𝟐 = −𝒓 _𝒅𝒕 − 𝑲𝒙 … … … … . . (𝟒)
𝒅^𝟐𝒙 𝒅𝒙
𝒎 _𝒅𝒕𝟐 + 𝒓 _𝒅𝒕 + 𝑲𝒙 = 𝟎 … … … … (𝟓)
Now, divide entire equation by mass(m)
^𝒅𝟐^𝒙 + ^{𝒓 𝒅𝒙} + ^𝒌 𝒙 = 𝟎……….(6)
𝒅𝒕^𝟐 𝒎 𝒅𝒕 𝒎
This equation represents differential equation of damped oscillation We know that,

now, the equation (6) become

r/m = 2b = damping co-efficient k/m = 𝝎^𝟐 = natural frequency

^𝒅𝟐^𝒙 + 𝟐𝒃 ^𝒅𝒙 + 𝝎^𝟐 𝒙 = 𝟎…………..(7)

𝒅𝒕^𝟐
𝒅𝒕

Let the solution of the above equation be ,
𝒙 = 𝑨е^𝜶𝒕 (𝟖)
Where, A and α are constants Differentiate equation(8) with respect to time ‘t’
^𝒅𝒙 = 𝑨𝜶е^𝜶𝒕 (9)
𝒅𝒕
Differentiate equation(9) with respect to time ‘t’
^𝒅𝟐^𝒙 = 𝑨𝜶^𝟐е^𝜶𝒕 (10)
𝒅𝒕^𝟐
Substituting equation (10),(9)and (8) in equation(7)
𝑨𝜶^𝟐е^𝜶𝒕 + 𝟐𝒃 𝑨𝜶е^𝜶𝒕 + 𝝎^𝟐 𝑨е^𝜶𝒕 = 𝟎
𝑨е^𝜶𝒕 (𝜶^𝟐 + 𝟐𝒃 𝜶 + 𝝎^𝟐 ) = 𝟎 (11)
𝒙(𝜶^𝟐 + 𝟐𝒃 𝜶 + 𝝎^𝟐 ) = 𝟎 (using equation 8)

From above equation either x=0 or 𝜶^𝟐 + 𝟐𝒃 𝜶 + 𝝎^𝟐 =0 The x is not equal to zero hence,
𝜶^𝟐 + 𝟐𝒃 𝜶 + 𝝎^𝟐=0
The solution for above quadratic equation is given by
𝜶 = −𝒃 ± √(𝒃^𝟐 − 𝝎^𝟐) …………..(11)
Substituting equation 11 in general solution of damped oscillation becomes
_{𝒙 = 𝑪𝒆}(−𝒃+^√(𝒃^𝟐− 𝝎^𝟐) )𝒕 _{+ 𝑫𝒆}(−𝒃−^√(𝒃^𝟐− 𝝎^𝟐) )𝒕_{………(12)} Where C and D are constant
^{Let time be maximum and the displacement should be maximum of x}o
^{Where, t=0 and x=x}o
Then the equation 14 becomes
^Xo^{= C + D (13)}

At maximum displacement the velocity, ^𝒅𝒙 = 𝟎

𝒅𝒕
Differentiate equation(12) and equate velocity to zero
^𝒅𝒙 = (−𝒃 + ^√(𝒃^𝟐 − 𝝎^𝟐)) 𝑪𝒆^{(−𝒃+√(𝒃}𝟐^{− 𝝎}𝟐^{) )𝒕} + (−𝒃 − ^√(𝒃^𝟐 − 𝝎^𝟐) ) 𝑫𝒆^{(−𝒃−√(𝒃}𝟐^{− 𝝎}𝟐^{) )𝒕}
𝒅𝒕
= 𝟎
Since t=0 when ^𝒅𝒙 = 𝟎
𝒅𝒕
The above equation become
(−𝒃 + ^√(𝒃^𝟐 − 𝝎^𝟐)) 𝑪 + (−𝒃 − ^√(𝒃^𝟐 − 𝝎^𝟐) ) 𝑫 = 𝟎
By rearrangement of equation
-b(C+D) +√(𝒃^𝟐 − 𝝎^𝟐) (C-D) =0
^-bxo ^{+ √(𝒃𝟐 − 𝝎𝟐) (C-D) =0 (x}o = ^C+D)

C – D = ^{𝐛𝐱𝐨
√}(𝒃^𝟐− 𝝎^𝟐)

…………………….(14)

By ^{adding equation (13) and (14)
2C = x}o ^{[𝟏 + 𝐛 ]
√}(𝒃^𝟐− 𝝎^𝟐)

^{C = x}o^{/2 [𝟏 + 𝐛
√}(𝒃^𝟐− 𝝎^𝟐)
^] …………….(15)

Similarly subtract equation (13) and (14)
^{2D = x}o ^{[𝟏 − 𝐛 ]
√}(𝒃^𝟐− 𝝎^𝟐)

^{D = x}o^{/2 [𝟏 + 𝐛
√}(𝒃^𝟐− 𝝎^𝟐)
] …………(16)

Substitute value of C and D in equation (11)

𝒙 = ^𝒙𝒐 [ [𝟏 + ^𝐛
_{] 𝒆}(−𝒃+^√(𝒃^𝟐− 𝝎^𝟐) )𝒕 _{+ [𝟏 +} ^𝐛
_{] 𝒆}(−𝒃−^√(𝒃^𝟐− 𝝎^𝟐) )𝒕 _]……(17)

^{𝟐 √}(𝒃^𝟐− 𝝎^𝟐) ^√(𝒃^𝟐− 𝝎^𝟐)
The above equation represents the displacement of a damped oscillation Types of damped oscillation

• 
  Over damping or dead beat case ⁽𝒃^𝟐 > 𝒘^𝟐)
  
• 
  Critical damping case ⁽𝒃^𝟐 = 𝒘^𝟐)
  
• 
  Under damped case ⁽𝒃^𝟐 < 𝒘^𝟐)
  

On the bases of equation (17) we can classify the damped oscillation in to above three types
Over damping : It is the condition under which the restoring and resistive forces acting on a body ar such that the body bought to halt at equilibrium position without oscillation due
to ⁽𝒃^𝟐 > 𝒘^𝟐) in equation (17) such a case is called as over damping case
Critical damping: It is the condition under which the restoring and resistive forces acting on a body are such that, the body is bought to halt at equilibrium without oscillation in
minimum time due to ⁽𝒃^𝟐 = 𝒘^𝟐) in equation(17) such a case is called as critical damping
case
Under damping: It is the condition under which the restoring and resistive forces acting on body are such that the body vibrates with diminishing of amplitude as the time
progresses and ultimately comes to halt at equilibrium position due to ⁽𝒃^𝟐 < 𝒘^𝟐) in
equation (17) such a case is called as under damping case

DERIVE AN EXPRESSION FOR AMPLITUDE AND PHASE OF A FORCED OSCILLATION

The forced oscillation system is subjected to

1. 
   A restoring force which is proportional to displacement but oppositely directed. This is written as -Kx, where µ is a constant of proportionality or force constant.
   
2. 
   A resistive force proportional to velocity but oppositely directed. This may be written as
   

- r , where r is the frictional force per unit velocity.

3. 
   an external force FSinpt is applied on damping system to reoscillate a body where p-angular frequency of external force
   

^{Resistive Force ( F}R^{) = −𝒓 𝒅𝒙 (𝟏)}
𝒅𝒕
^{Restoring force (F}x^{) = -Kx (2)
External force (F}ext^{) = Fsinpt (3)}
By using newton’s 2^nd law
𝑭 = 𝒎 ^𝒅𝒙 … … … … … … … (𝟒)
𝒅𝒕
Now total force acting on a body is given by
F = RESISTIVE FORCE + RESTORING FORCE + EXTERNAL FORCE
Using equation (1), (2) ,(3) and (4)

𝒅^𝟐𝒙
𝒅𝒙

𝒎 _𝒅𝒕𝟐 = −𝒓 _𝒅𝒕 − 𝑲𝒙 + 𝑭𝒔𝒊𝒏𝒑𝒕 … … … … . . (𝟒)
𝒅^𝟐𝒙 𝒅𝒙
𝒎 _𝒅𝒕𝟐 + 𝒓 _𝒅𝒕 + 𝑲𝒙 = 𝑭𝒔𝒊𝒏𝒑𝒕 … … … … (𝟓)
Now, divide entire equation by mass(m)
^𝒅𝟐^𝒙 + ^{𝒓 𝒅𝒙} + ^𝒌 𝒙 = ^𝒇 𝒔𝒊𝒏𝒑𝒕……….(6)
𝒅𝒕^𝟐 𝒎 𝒅𝒕 𝒎 𝒎
This equation represents differential equation of forced oscillation We know that,

now, the equation (6) become

r/m = 2b = damping co-efficient k/m = 𝝎^𝟐 = natural frequency

^𝒅𝒙 + 𝟐𝒃 ^𝒅𝒙 + 𝝎^𝟐 𝒙 = ^𝒇 𝒔𝒊𝒏𝒑𝒕…………..(7)
𝒅𝒕 𝒅𝒕 𝒎
Let the solution of the above equation be ,
𝒙 = 𝐚𝐬𝐢𝐧(𝒑𝒕 − 𝜶) … … … … . . (𝟖)

Where, a-amplitude and α –phase of forced oscillation Differentiate equation(8) with respect to time ‘t’
^𝒅𝒙 = 𝐚𝐩𝐂𝐨𝐬(𝒑𝒕 − 𝜶)……………(9)
𝒅𝒕
Differentiate equation(9) with respect to time ‘t’
^𝒅𝟐^𝒙 = − 𝐚𝐩^𝟐𝐬𝐢𝐧(𝒑𝒕 − 𝜶)………………(10)
𝒅𝒕^𝟐
Substituting equation (10),(9)and (8) in equation(7)
− 𝐚𝐩^𝟐𝐬𝐢𝐧(𝒑𝒕 − 𝜶) + 𝟐𝒃𝐚𝐩𝐂𝐨𝐬(𝒑𝒕 − 𝜶) + 𝝎^𝟐 𝐚𝐬𝐢𝐧(𝒑𝒕 − 𝜶) = ^𝒇 𝒔𝒊𝒏𝒑𝒕 ………..(11)
𝒎
Consider only Left hand side of the equation only
^𝒇 𝒔𝒊𝒏𝒑𝒕 = ^𝒇 𝐬𝐢𝐧^[(𝒑𝒕 − 𝜶⁾ + 𝜶^] = ^𝒇 [𝐬𝐢𝐧⁽𝒑𝒕 − 𝜶⁾𝒄𝒐𝒔𝜶 + 𝒔𝒊𝒏𝜶 𝐜𝐨𝐬(𝒑𝒕 − 𝜶)…….(12)
𝒎 𝒎 𝒎
Apply equation (12) in equation (11) Then,
− 𝐚𝐩^𝟐𝐬𝐢𝐧(𝒑𝒕 − 𝜶) + 𝟐𝒃𝐚𝐩𝐂𝐨𝐬(𝒑𝒕 − 𝜶) + 𝝎^𝟐 𝐚𝐬𝐢𝐧(𝒑𝒕 − 𝜶) = ^𝒇 [𝐬𝐢𝐧⁽𝒑𝒕 − 𝜶⁾𝒄𝒐𝒔𝜶 +
𝒎
𝒔𝒊𝒏𝜶 𝐜𝐨𝐬(𝒑𝒕 − 𝜶)] ……………(13)
Equate the co-efficient of 𝐬𝐢𝐧(𝒑𝒕 − 𝜶) by the equation(13)

− 𝐚𝐩^𝟐+𝝎^𝟐 𝐚 = ^𝒇
𝒎
a(𝝎^𝟐 − 𝐩^𝟐) = ^𝒇
𝒎
𝒄𝒐𝒔𝜶
𝒄𝒐𝒔𝜶 …………….(14)

similarly equate the co-efficient of 𝐂𝐨𝐬(𝒑𝒕 − 𝜶) by the equation(13)
𝟐𝒃𝐚𝐩 = ^𝒇 𝒔𝒊𝒏𝜶……………….(15)
𝒎
Squaring and add equation(14) and (15)

𝒂^𝟐(𝝎^𝟐 − 𝒑^𝟐)𝟐 + 𝟒𝒃^𝟐𝒂^𝟐𝒑^𝟐 = (

_𝒇 𝟐
)


[𝐬𝐢𝐧^𝟐 𝜶 + 𝐜𝐨𝐬^𝟐 𝜶]

𝒎



+ 𝟒𝒃 𝒑 ] = ( )
𝒂^𝟐[⁽𝝎^𝟐 − 𝒑^{𝟐)𝟐 𝟐 𝟐 𝒇} 𝟐
𝒎

𝒂^𝟐 =

𝒇 ^𝟐
(_𝒎)

^[(𝝎^𝟐 − 𝒑^𝟐)𝟐 + 𝟒𝒃^𝟐𝒑^𝟐]
Apply square root on both side

The above equation represents the amplitude of forced vibration Now, divide equation (15) and (14)

𝟐𝒃𝐚𝐩/𝐚(𝝎^𝟐 − 𝐩^𝟐) = ^𝒇
𝒎
𝒄𝒐𝒔𝜶/ ^𝒇 𝒔𝒊𝒏𝜶
𝒎

𝜶 = 𝐭𝐚𝐧^−𝟏( 𝟐𝒃𝐩/(𝝎^𝟐 − 𝐩^𝟐))

𝒕𝒂𝒏𝜶 = 𝟐𝒃𝐩/(𝝎^𝟐 − 𝐩^𝟐)
The above equation represents the phase of the forced oscillation
RESONANCE:
If we bring a vibrating tuning fork near another stationary tuning fork of the same natural frequency as that of vibrating tuning fork, we find that stationary tuning fork also starts vibrating. This phenomenon is known as resonance. The phenomenon of making a body vibrate with its natural frequency under the influence of another vibrating body with the same frequency is called resonance.
CONDITION FOR RESONANCE

1. 
   the damping by a system should be minimum( b valueshould be less)
   
2. 
   the amplitude must be maximum ( p = 𝝎)
   

then, amplitude becomes

𝒇

𝑨𝒎𝒂𝒙 = ^𝒎
𝟐𝒃𝒑


EXAMPLES FOR RESONANCE:

1. 
   Oscillations of the stretched string kept under the influence of oscillating magnetic field caused by oscillating current. The string vibrates with maximum amplitude when the applied frequency matches with Natural frequency of the string.
   
2. 
   Sodium chloride crystal has alternately Sodium and Chloride ions. If an electric field is applied on the crystal, the charges would oscillate back and forth. The natural frequency is in Infrared range.
   
3. 
   Helmholtz resonator
   
4. 
   tuning of radio
   

5. 
   tuning of string instruments
   

EXAMPLES FOR DAMPED OSCILLATION

1. 
   mechanical oscillations of simple pendulum
   
2. 
   Electrical oscillations of LC circuit in which the circuit is closed just the capacitor is charged
   
3. 
   A swing free to oscillate after being pushed once EXAMPLES FOR FORCED OSCILLATION
   
   1. 
      Oscillation of a swing which is pushed periodically by a person
      
   2. 
      The periodic variation of current in an LCR circuit driver by an AC source
      
   3. 
      The vibrations of ear drum caused by sound from a sounding body (such as a tuning fork)
      
   4. 
      The motion of hammer in a calling bell
      
   5. 
      The motion of diaphragm in a telephone receiver or a loud speaker
      

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Only for Sāmbhram university students, jizzax , Uzbekistan