Forced vibrations/oscillations
When an external periodic force acts on an oscillating body, the body is forced to
oscillate with the frequency of applied force. This kind of oscillation is called forced
oscillation.
Theory of forced vibrations/oscillations
Consider a body of mass m displaced through a distance x at any instant of time t in a
resistive medium. Let an external periodic force F sin(pt) act on it in the opposite
direction. Where F is the magnitude of applied periodic force and p is its angular
frequency.
The resultant force acting on the body is given by the sum of resistive force, restoring
force and external periodic force.
Therefore, the resultant force acting on the body is given by
where r is damping constant and k is force constant
According to Newton’s second law of motion, the resultant force on the body is
From equations 1 and 2, we get
This is equation of motion for forced oscillation
On re-arranging the above equation, we get
The solution of this differential equation is x =𝑎 𝑆𝑖𝑛(𝑝𝑡−𝛼) ……(4)
Where a and 𝛼 represent amplitude and phase of the oscillating body.
Differentiating equation 4, w.r.t ‘ t ‘ twice ,we get
Substituting in equation 5 and simplifying, we get
Equating the coefficients of 𝑆𝑖𝑛(𝑝𝑡−𝛼) and 𝐶𝑜𝑠(𝑝𝑡−𝛼) on both sides separately we get
−𝑎𝑝2+𝜔2 𝑎 = 𝐹/𝑚 𝐶𝑜𝑠𝛼 …(6)
2𝑏 𝑎𝑝= 𝐹/𝑚 𝑆𝑖𝑛𝛼 ….(7)
Squaring and adding equations 6&7, we get
This is the equation for amplitude of the forced vibrations.
From equations 4&8 , we get
The phase 𝛼 of the forced vibration is obtained by dividing equations 7 by 6,
Dependence of amplitude (a) and phase(𝜶) on the frequency (p) of
applied force:
We know that, amplitude is given by
and Phase is given by
Case(1):
As 𝑝2 is very small, then 𝜔2−𝑝2≈𝜔2, 2bp=0 & 2𝑏𝑝/𝜔2≈0
∴ from equation 1, amplitude is
Thus a is independent of p but depends on (F/m) and constant for given F.
Also from equation 2, 𝛼=𝑡𝑎𝑛−1[𝟎]=0,thus displacement and force will be in same phase.
Case(2): For p=𝝎, 𝜔2−𝑝2=0, 2bp=0
∴ From equation 1, amplitude,
Thus a will have highest value for a given damping force F.
And phase, 𝛼=𝑡𝑎𝑛−1[2𝑏𝑝/0]= =𝑡𝑎𝑛−1[∞]= 𝜋/2 ,thus displacement phase lags 𝜋/2 with
respect to phase of applied force.
Case(3): For p ≫𝝎 is applicable for small b, (𝜔2−𝑝2)2≈(𝑝2)2=𝑝4
∴� From equation 1, amplitude is
but for small b
thus as p increases, a decreases
Also phase, 𝛼=𝑡𝑎𝑛−1[𝟐𝒃𝒑/(𝝎𝟐−𝒑𝟐)] =𝑡𝑎𝑛−1[𝟐𝒃𝒑/−𝒑𝟐] =𝑡𝑎𝑛−1[𝟐𝒃/−𝒑] but for small b,
2𝑏𝑝≈0
∴ 𝛼=𝑡𝑎𝑛−1[−𝟎] = 𝜋 , thus for large p ,the displacement phase lags by 𝜋 w.r.t phase of
applied force.
Resonance:
When the frequency of the external periodic force acting on the body becomes equal to
the natural frequency of the body, the situation is called resonance. At resonance, the
energy transfer from external periodic force is maximum and the amplitude of oscillation
also becomes maximum.
Conditions for resonance:
1. The frequency of the applied force (p) must be equal to the natural frequency (𝜔) of
oscillations of the body.
2. b = 𝑟/2𝑚 must be minimum or Damping caused by the medium must be minimum.
At resonance the amplitude is given by
The amplitude of the body near resonance
Sharpness of Resonance:
The sharpness of resonance is the ratio of change in amplitude (Δ𝑎 ) to corresponding
change in frequency(Δ𝜔 ) of the applied external periodic force, at resonance.
ie: Sharpness of resonance = Δ𝑎/Δ𝜔
Effect of damping on sharpness of resonance:
The variation of amplitude of forced oscillations with respect to damping is as shown in
the graph.
From the graph it is clear that the maximum amplitude at resonance is a function of
damping. Higher the damping lower will be the amplitude at resonance. Thus the
sharpness will be higher at lower damping and vice-versa.
Example of resonance:
1. Helmholtz resonator (HR)
2. Tuning of radio receiver set to the broadcasting transmitting frequency.
3. Setting up of standing waves in Melde’s experiment string.
Helmholtz resonator (HR)
Helmholtz resonator is a device used to detect the presence of sound of a particular
frequency in the mixture of sound of different frequencies. It is also used to tune an
instrument to the given note.
Construction:
It consists of a hollow metallic sphere with a long cylindrical neck (A) and a fine hole (B)
opposite to A. The air inside the sphere has a definite natural frequency which is marked
on it.
Working:
When sound of different frequencies enters the resonator through A, the air inside the
resonator resonates for the frequency of the sound which is equal to its natural frequency.
This matching of frequency produces resonance, which can be heard as loud sound at the
end B. Helmholtz resonator cannot resonate for any other frequency.
