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## Forced Oscillation and Resonance, Physics tutorial

:Differential equation for a weakly damped forced oscillatorTo 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 F

_{0}is the constant:F(t) = F

_{0}cosωt.......................................Eq.1Let 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) = F

_{0}cosωt.So for forced oscillator Equation is altered to

md

^{2}x/dt^{2}= -kx-γdx/dt + F_{0}cosωt.......................................Eq.2Dividing by m and rearranging terms, equation of motion of the forced oscillator takes form

d

^{2}x/dt^{2}+ 2bdx/dt + ω_{0}^{2}x = f_{0}cosωt.......................................Eq.3Where 2b = γ/m, ω

_{0}^{2}= k/m and f_{0}= F_{0}/m is the measure of driving force.:Solutions of the differential equationWhen there is no applied force, weakly damped system (b < ω

_{0}) oscillates harmonically with angular frequency ω_{d}= √ω_{0}^{2}- b^{2}. 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 ω_{d}(of damped oscillations) and other of frequency ω(of driving force). Therefore when ω ≠ ω general solution can be written as:x(t) = x

_{1}(t) + x_{2}(t)Where x

_{1}(t) is the solution of equation obtained by replacing RHS of Eq.3 by zero. On substituting result in Eq.3, then x_{2}(t) satisfies equation:d

^{2}x_{2}/dt^{2}+ 2bdx_{2}/dt + ω_{0}^{2}x^{2}= f_{0}cosωtWhen there is no driving force, displacement of the weakly damped (b < ω

_{0}) system at any instant is:x

_{1}(t) = a_{0}e^{-bt}cos(ω_{d}t + 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.

Steady-state Solution:To get steady state solution of Eq.3, let us assume that displacement of forced oscillator is given by

x

_{2}(t) = -aωcos(ωt - θ) .......................................Eq.4Where 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,

dx

_{2}/dt = -aωsin(ωt - θ) and d^{2}x_{2}/dt^{2}= -aω^{2}cos(ωt - θ)Substituting the results back in Eq.3, we get

(ω

_{0}^{2 }- ω^{2}) acos(ωt - θ)- 2abωsin(ωt - θ) = f_{0}cosωtOn rearranging terms equation will be:

[(ω

_{0}^{2}- ω^{2})acosθ + 2abωsinθ - f_{0}]cosωt + [(ω_{0}^{2}- ω^{2})asinθ - 2abωcos]sinωt = 0.......................................Eq.5We 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.

(ω

_{0}^{2}- ω^{2})acosθ + 2abωsinθ = f_{0}.......................................Eq.6(a)(ω

_{0}^{2}- ω^{2})asinθ - 2abωcos = 0.......................................Eq.6(b)Eq.6(b) readily provides phase by which driving force leads displacement:

θ = tan

^{-1}2bω/(ω_{0}^{2}- ω^{2})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) = a

_{0}e^{-bt}cos(ω_{d}t + Φ) + F_{0}cos(ωt - θ)/m[(ω_{0}^{2}- ω^{2})^{2}+ 4b^{2}ω^{2}]^{1/2}For undamped system, steady-state solution is:

θ = 0 and x

_{2}(t) = F_{0}/m(ω_{0}^{2}- ω^{2})cosωtThat 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 oscillationsDepending on relative magnitudes of natural and driving frequencies, three cases occur.

i)

Low Driving Frequency (ω << ω:_{0})To know behavior of a(ω) at low driving frequencies,

a(ω) = f

_{0}/ω_{0}^{2}[(1 - ω^{2}/ω_{0}^{2})^{2}+ 4b^{2}ω^{2}/ω_{0}^{2}]^{1/2}For ω << ω

_{0}ratio ω_{2}/ω_{0}^{2}will be much less than one. So, we neglect terms having ω^{2}/ω_{0}^{2}. This providesa(ω) = f

_{0}/ω_{0}^{2}= F_{0}/mω_{0}^{2}= F_{0}/kTherefore, 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/(ω

_{0}^{2}- ω^{2}) → 0 for ω/ω_{0}<< 1That 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 = ω

_{0}. First term in denominator vanishes and amplitude is provided bya(ω

_{0}) = f_{0}/2bω_{0}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 = ω

_{0},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:

a

_{max}= f_{o}/2b√ω_{0}^{2}- b^{2}When at the particular frequency, amplitude of driven system becomes maximum; we say that amplitude resonance takes place. Frequency ω

_{r}is referred to as resonance frequency. It is instructive to note that ω_{r}is less than ω_{0}as well as ω_{d}= √ω_{0}^{2}- b^{2}.iii)

High Driving Frequency:For ω >> ω

_{0}a(ω) = f

_{0}/ω^{2}[(1 - ω_{0}^{2}/ω^{2})^{2}+ 4b^{2}/ω^{2}]^{1/2}Then amplitude of resulting vibration is given by

a(ω) = f

^{0}/ω^{2}That is, at high frequencies amplitude decreases as 1/ω

^{2}and eventually becomes zero.Similarly phase is provided by

tanθ = 2bω/(ω

_{0}^{2}- ω^{2}) ≈ -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 ω

_{r}= √(ω_{0}^{2}- 2b^{2}) and has value f_{0}/2b√(ω_{0}^{2}- b^{2}). For ω > ω_{r}, 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 = ω

_{0}, θ = π/2.:Power absorbed by a forced oscillatorEvery 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 = -v

_{0}sin(ωt - θ) = v_{0}cos(ωt - Φ)Where v

_{0}= f_{0}ω/m[(ω_{0}^{2}- ω^{2})^{2}+ 4b^{2}ω^{2}]^{1/2}is velocity amplitude and Φ = θ - π/2It is the phase difference between velocity and applied force. Instantaneous power absorbed by oscillator is given by

P(t) = F

_{0}v_{0}cosωtcos(ωt - Φ)Average power absorbed by forced oscillator will be maximum when sinθ = 1 = cosΦ , i.e., Φ = π/2 (ω = ω

_{0}). This happens for ω = ω_{0}<P>max = 1/4bmF

_{0}^{2}That is, peak value of average power absorbed by the maintained system is determined by damping and amplitude of driving force.

:Quality factorThe 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)(F

_{0}ω_{0}/kQ)(ω_{0}^{2}ω^{2})/[(ω_{0}^{2}- ω^{2})^{2}+ ω_{0}^{2}ω^{2}/Q^{2}] = (1/4)(F_{0}^{2}ω_{0}Q/k)On simplification the equation is:

(ω

_{0}^{2}- ω^{2}) = ω^{2}ω_{0}^{2}/Q^{2}so that (ω

_{0}^{2}- ω^{2}) = ±ω^{2}ω_{0}^{2}/Q^{2}This equation has 4 roots. Of these, two roots correspond to negative frequencies and are physically unacceptable. The other two acceptable roots are

ω

_{1}= -ω_{0}/2Q + ω_{0}(1 + 1/4Q^{2})1/2ω

_{2}= ω_{0}/2Q + ω_{0}(1 + 1/4Q^{2})1/2Frequency interval between two half-power points is

ω

_{2}- ω_{1}= 2Δω = ω_{0}/QIt 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.

:LCR CIRCUITThe 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 = E

_{0}cosωtAs I = dq/dt equation can be written as

Ld

^{2}q/dt^{2}+ Rdq/dt + q/C = E_{0}cosωtDividing through by L, we get

d

^{2}q/dt^{2}+ (R/L)dq/dt + q/LC = (E_{0}/L)cosωtFor the weakly damped system, charge on capacitor plates at any instant of time is given by

q = ((E

_{0}/L)/[(1/LC - ω^{2}) + (ωR/L)^{2}]^{1/2}) cos(ωt - θ)Where ω = √1/LC - R

^{2}/4L^{2}is angular frequency of oscillation and tan θ = (ωR/L)/(1/LC - ω^{2}) states phase with respect to applied emf.Current in circuit is attained by differentiating with respect to t. The result is

I = [E

_{0}√(R^{2}+ (ωL - 1/ωC)^{2})]cos(ωt - Φ)Where Φ = θ - π/2 is the phase difference between E

_{0}and I. SincetanΦ = -cotθ = (1/LC - ω

^{2})/(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 I

_{0}/E_{0}/R and circuit is said to resonate with frequencyvr = 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 = ω

_{0}L/RWhere ω

_{0}= 1/√LCYou can verify that bandwidth of power resonance curve for LCR circuit is provided by

ω

_{2}- ω_{1}= 2/τ = R/L = ω_{0}/QSo 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 10

^{2}to 10^{3}. Microwave cavities have Q values of order of 10^{5}.Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)Expand your confidence, grow study skills and improve your grades.

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