Differential equation for a weakly damped forced oscillator:
To set up differential equation of the forced weakly damped harmonic oscillator, consider spring-mass system. It is now also subjected to the external driving force, F(t). That is, instead of allowing model oscillator to oscillate at natural frequency, we push it backward and forward periodically at the frequency ω. We can write driving force as where F0 is the constant:
F(t) = F0cosωt.......................................Eq.1
Let mass be displaced from equilibrium position and then released. At any instant, it is subject to (i) restoring force, -kx , (ii) damping force, -γdx/dt and (iii) driving force, F(t) = F0cosωt.
So for forced oscillator Equation is altered to
md2x/dt2 = -kx-γdx/dt + F0cosωt.......................................Eq.2
Dividing by m and rearranging terms, equation of motion of the forced oscillator takes form
d2x/dt2 + 2bdx/dt + ω02x = f0cosωt.......................................Eq.3
Where 2b = γ/m, ω02 = k/m and f0 = F0/m is the measure of driving force.
Solutions of the differential equation:
When there is no applied force, weakly damped system (b < ω0) oscillates harmonically with angular frequency ωd = √ω02 - b2. But when the driving force of angular frequency ω is applied, it enforces its own frequency on oscillator. Therefore, we expect that actual motion will be the result of superposition of two oscillations; one of frequency ωd (of damped oscillations) and other of frequency ω(of driving force). Therefore when ω ≠ ω general solution can be written as:
x(t) = x1(t) + x2(t)
Where x1(t) is the solution of equation obtained by replacing RHS of Eq.3 by zero. On substituting result in Eq.3, then x2(t) satisfies equation:
d2x2/dt2 + 2bdx2/dt + ω02x2 = f0cosωt
When there is no driving force, displacement of the weakly damped (b < ω0) system at any instant is:
x1(t) = a0e-btcos(ωdt + Φ)
This complementary function decays exponentially and after some time it will disappear. That is why it is also referred to as transient solution. In transient state, system oscillates with some frequency which is other than natural frequency or frequency of driving force.
To get steady state solution of Eq.3, let us assume that displacement of forced oscillator is given by
x2(t) = -aωcos(ωt - θ) .......................................Eq.4
Where a and θ are unknown constants. By comparing Eqs.1 and 4 driving force leads displacement in phase by angle θ.
To determine a and θ we differentiate Eq.4 twice with respect to time. This gives,
dx2/dt = -aωsin(ωt - θ) and d2x2/dt2 = -aω2cos(ωt - θ)
Substituting the results back in Eq.3, we get
(ω02 - ω2) acos(ωt - θ)- 2abωsin(ωt - θ) = f0cosωt
On rearranging terms equation will be:
[(ω02 - ω2)acosθ + 2abωsinθ - f0]cosωt + [(ω02 - ω2)asinθ - 2abωcos]sinωt = 0.......................................Eq.5
We know that both cosωt and sinωt never concurrently become zero; when one vanishes, other takes maximum value. Thus, Eq.5 can be satisfied only when both terms within square brackets become zero separately, i.e.
(ω02 - ω2)acosθ + 2abωsinθ = f0.......................................Eq.6(a)
(ω02 - ω2)asinθ - 2abωcos = 0.......................................Eq.6(b)
Eq.6(b) readily provides phase by which driving force leads displacement:
θ = tan-12bω/(ω02 - ω2)
Steady-state solution has frequency of driving force and its amplitude is constant. Furthermore, phase is also stated completely with respect to driving force. Thus, it doesn't depend on initial conditions. In other words, motion of the driven system in steady-state is independent of way we start oscillation. Transient solution, steady-state solution and their sum,
x(t) = a0e-btcos(ωdt + Φ) + F0cos(ωt - θ)/m[(ω02 - ω2)2 + 4b2ω2]1/2
For undamped system, steady-state solution is:
θ = 0 and x2(t) = F0/m(ω02 - ω2)cosωt
That is, driving force and displacement are in phase (θ = 0). Phase lag is fundamentally consequence of damping. If frequency of driving force equals frequency of undamped oscillator, amplitude will become infinitely large. Then resonance is said to occur.
Effect of the frequency of the driving force on the amplitude and phase of steady-state forced oscillations:
Depending on relative magnitudes of natural and driving frequencies, three cases occur.
i) Low Driving Frequency (ω << ω0):
To know behavior of a(ω) at low driving frequencies,
a(ω) = f0/ω02[(1 - ω2/ω02)2 + 4b2ω2/ω02]1/2
For ω << ω0 ratio ω2/ω02 will be much less than one. So, we neglect terms having ω2/ω02. This provides
a(ω) = f0/ω02 = F0/mω02 = F0/k
Therefore, at very low driving frequencies, steady-state amplitude of oscillation is controlled by stiffness constant and magnitude of driving force. Under this condition
tanθ = 2bω/(ω02 - ω2) → 0 for ω/ω0 << 1
That is, driving force and steady-state displacement are in phase.
ii) Resonance Frequency (ω = ω0):
To compute value of a(ω) at resonance, we set ω = ω0. First term in denominator vanishes and amplitude is provided by
a(ω0) = f0/2bω0
At resonance amplitude depends on damping; it is inversely proportional to b. That is why in actual practice amplitude never becomes infinite.
Similarly by setting ω = ω0,
tan θ → ∞ so that θ = π/2
This means that driving force and displacement are out of phase by π/2.
Te peak value of steady-state amplitude:
amax = fo/2b√ω02 - b2
When at the particular frequency, amplitude of driven system becomes maximum; we say that amplitude resonance takes place. Frequency ωr is referred to as resonance frequency. It is instructive to note that ωr is less than ω0 as well as ωd = √ω02 - b2.
iii) High Driving Frequency:
For ω >> ω0
a(ω) = f0/ω2[(1 - ω02/ω2)2 + 4b2/ω2]1/2
Then amplitude of resulting vibration is given by
a(ω) = f0/ω2
That is, at high frequencies amplitude decreases as 1/ω2 and eventually becomes zero.
Similarly phase is provided by
tanθ = 2bω/(ω02 - ω2) ≈ -2b/ω →0(ω→∞)
θ = π
This signifies that at high frequencies driving force and displacement are out of phase by π.
Therefore conclude that:
(i) Amplitude of oscillation in steady-state differs with frequency. It becomes maximum ωr = √(ω02 - 2b2) and has value f0/2b√(ω02 - b2). For ω > ωr, a(ω) decreases, as ω-2.
(ii) Displacement lags behind the driving force by angle θ, which increases from zero at ω = 0 to π at extremely high frequencies. At ω = ω0, θ = π/2.
Power absorbed by a forced oscillator:
Every oscillating system loses energy in doing work against damping. But oscillations of the forced oscillator are maintained by energy supplied by driving force. It is, thus, significant to know average rate at which energy should be supplied to system to maintain steady-state oscillations. Now compute average power absorbed by oscillating system. By definition, instantaneous power is given by
P(t) = force X velocity
= F(t) x v
v = -v0sin(ωt - θ) = v0cos(ωt - Φ)
Where v0 = f0ω/m[(ω02 - ω2)2 + 4b2ω2]1/2 is velocity amplitude and Φ = θ - π/2
It is the phase difference between velocity and applied force. Instantaneous power absorbed by oscillator is given by
P(t) = F0v0cosωtcos(ωt - Φ)
Average power absorbed by forced oscillator will be maximum when sinθ = 1 = cosΦ , i.e., Φ = π/2 (ω = ω0). This happens for ω = ω0
<P>max = 1/4bmF02
That is, peak value of average power absorbed by the maintained system is determined by damping and amplitude of driving force.
The quality factor of a damped oscillator defined as:
Q = 2πaverage energy stored in one cycle/average energy dissipated in one cycle
Use same definition to calculate Q of forced oscillator once you know <E> and <P>.
i) Q in Terms of Band Width: Sharpness of a Resonance:
The Q of a system can also be defined as
Q = (Frequency at which powerresonance occurs)/(Fullwidthat half powerpoints)
To calculate the frequency at which average power drops to half its maximum value we can write
(1/2)(F0ω0/kQ)(ω02ω2)/[(ω02 - ω2)2 + ω02ω2/Q2] = (1/4)(F02ω0Q/k)
On simplification the equation is:
(ω02 - ω2) = ω2ω02/Q2
so that (ω02 - ω2) = ±ω2ω02/Q2
This equation has 4 roots. Of these, two roots correspond to negative frequencies and are physically unacceptable. The other two acceptable roots are
ω1 = -ω0/2Q + ω0(1 + 1/4Q2)1/2
ω2 = ω0/2Q + ω0(1 + 1/4Q2)1/2
Frequency interval between two half-power points is
ω2 - ω1 = 2Δω = ω0/Q
It is clear that high Q system has small bandwidth and resonance is said to be sharp. On the other hand, a low Q system has large bandwidth and resonance is said to be flat. Therefore, sharpness of resonance refers to rapid rate of fall of power with frequency on either side of resonance. Q factor has its greatest significance in reference to electrical circuits.
The resonant behavior of the simple mechanical system subject to periodic force. Another physical system that also shows resonant behavior is series LCR circuit containing a source of alternating emf. We will discuss the behavior of this system by drawing similarities with a mechanical system.
In an LCR circuit, charge oscillations die out because of power losses in the resistance. Let I be current in the circuit at a given time. Then, applied emf is equal to the sum of the potential differences across the capacitor, resistor and the inductor. Then equation can be written as:
q/C + RI + LdI/dt = E0cosωt
As I = dq/dt equation can be written as
Ld2q/dt2 + Rdq/dt + q/C = E0cosωt
Dividing through by L, we get
d2q/dt2 + (R/L)dq/dt + q/LC = (E0/L)cosωt
For the weakly damped system, charge on capacitor plates at any instant of time is given by
q = ((E0/L)/[(1/LC - ω2) + (ωR/L)2]1/2) cos(ωt - θ)
Where ω = √1/LC - R2/4L2 is angular frequency of oscillation and tan θ = (ωR/L)/(1/LC - ω2) states phase with respect to applied emf.
Current in circuit is attained by differentiating with respect to t. The result is
I = [E0√(R2 + (ωL - 1/ωC)2)]cos(ωt - Φ)
Where Φ = θ - π/2 is the phase difference between E0 and I. Since
tanΦ = -cotθ = (1/LC - ω2)/(ωR/L)
As driving frequency increases, reactance (ωL - 1/ωC) decreases and current amplitude increases. When ωL = 1/ωC. Term under radical sign in becomes minimum; equal to R. Then current obtains its peak value I0/E0/R and circuit is said to resonate with frequency
vr = 1/(2π√LC)
At resonance, current and applied emf are in phase. When driving frequency is high, circuit will be inductive and current lags behind emf by π/2.
The quality factor of an LCR circuit is given by
A = ω0L/R
Where ω0 = 1/√LC
You can verify that bandwidth of power resonance curve for LCR circuit is provided by
ω2 - ω1 = 2/τ = R/L = ω0/Q
So that Q = (Frequency at resonance) / (Full width at half - power points)
Q of circuit finds its ability to choose narrow band of frequencies from the wide range of input frequencies. This, hence, obtains particular significance in relation to radio receivers. Signals of different frequencies from all stations are present around antenna. But receiver selects just one particular station to which we wish to tune and discards others. Usually, radio receivers operating in MHz region have Q values of order of 102 to 103. Microwave cavities have Q values of order of 105.
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