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Waves and Oscillations
Forced Oscillations and Resonance
Forced oscillations occur when a mechanical system is subjected to a continuous periodic driving force. The system ultimately oscillates at the frequency of the external driver rather than its own natural frequency.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of forced oscillations, qualitative understanding of resonance, amplitude and frequency relations, and basic examples.
PGT extension
Detailed mathematical derivation of the steady-state amplitude and phase angle, sharpness of resonance, quality factor (Q-factor), and power absorption.
Prerequisites
Damped harmonic motion and damping coefficient
Differential equations of second order with constant coefficients
Complex numbers and phasor addition
Differential Equation of Forced Oscillations
A system of mass m subject to restoring force -kx, damping force -rv, and external periodic force F_0 cos(omega t) is governed by m(d^2x/dt^2) + r(dx/dt) + k$x = F_0 \cos(\omega t)$.
Dividing by m yields d^2x/dt^2 + 2gamma(dx/dt) + omega_0^2 $x = (F_0/m) \cos(\omega t)$, where gamm$a = r/(2m)$ is the damping factor and omeg$a_0 = \sqrt(k/m)$ is the natural angular frequency.
The complete solution consists of a transient part (which dies out due to damping) and a steady-state part (which persists at the driving frequency omega).
Steady-State Response and Phase Lag
In the steady state, the displacement is given by x(t) = A cos(omega t - phi), where A is the steady-state amplitude and phi is the phase lag of the displacement behind the driving force.
The phase angle phi is determined by tan(phi) = 2gamma omega / (omega_0^2 - omega^2).
At low driving frequencies (omega << omega_0), phi approx 0. At resonance (omeg$a = \omega_0)$, ph$i = \pi/2$. At high frequencies (omega >> omega_0), phi approaches pi.
Amplitude Resonance
The amplitude of forced oscillations is $A = (F_0/m) / \sqrt((\omega_0^2 - \omega^2)^2 + 4gamma^2 \omega^2)$.
Amplitude resonance occurs when the denominator is minimized with respect to omega, giving the resonant angular frequency omeg$a_r = \sqrt(\omega_0^2 - 2gamma^2)$.
For very light damping (gamma approx 0), the resonance frequency coincides with the natural frequency omega_0.
Quality Factor and Sharpness of Resonance
The Quality factor (Q) measures the sharpness of resonance and is defined as $Q = \omega_r / (2gamma) \approx \omega_0 / (\omega_2 - \omega_1)$, where omega_2 - omega_1 is the full width at half maximum of the resonance power curve.
A high Q-factor implies very light damping, a narrow resonance peak, and a large amplitude at resonance.
At resonance, the maximum amplitude is $A_max = F_0 / (2m \gamma \omega_r) = Q(F_0 / k)$.
Derivations
Steady-State Amplitude and Phase Angle Derivation
Conditions: Assumes steady-state solution x(t) = A cos(omega t - phi) for the equation d^2x/dt^2 + 2gamma(dx/dt) + omega_0^2 $x = (F_0/m) \cos(\omega t)$.
Substitute $x = A \cos(\omega t - \phi)$, dx/d$t = -A \omega \sin(\omega t - \phi)$, and $d^2x/dt^2 = -A \omega^2 \cos(\omega t - \phi) into the differential equation$.
Expand the trigonometric terms: cos(omega t - phi) = cos(omega t)cos(phi) + sin(omega t)sin(phi) and sin(omega t - phi) = sin(omega t)cos(phi) - cos(omega t)sin(phi).
Equate the coefficients of cos(omega t) and sin(omega t) on both sides of the equation to obtain two simultaneous equations for amplitude A and phase phi.
Square and add the two resulting equations to eliminate phi and solve for A, and divide them to find tan(phi).
Conditions: Finding the driving frequency omega that maximizes the steady-state amplitude A.
Express the amplitude A as a function of omega and differentiate with respect to omega to find extrema.
Equate the derivative dA/domega to zero, which is equivalent to minimizing the radicand function f(omega) = (omega_0^2 - omega^2)^2 + 4gamma^2 omega^2.
Differentiate f(omega) with respect to omega: df/domeg$a = 2(\omega_0^2 - \omega^2)(-2omega) + 8gamma^2 \omega = 0$.
$$\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega t)$$
x: displacement, gamma: damping coefficient, omega_0: natural frequency, F_0: driving force amplitude, omega: driving frequency m, s^-1, rad/s, N, rad/s
Linear damped oscillator with sinusoidal driving force
Draw: A family of curves plotting amplitude A on the y-axis against driving frequency omega on the x-axis for different values of damping gamma.
Interpret: As damping gamma decreases, the resonance peak becomes higher and sharper. For very large damping, the peak shifts towards zero and becomes flattened.
Exam use: Directly tested in conceptual MCQs regarding the effect of damping on peak height and resonant frequency location.
Phase Lag vs Driving Frequency Curve
Draw: A sigmoid-shaped curve plotting phase phi (from 0 to pi radians) against driving frequency omega, passing through pi/2 at omeg$a = \omega_0$.
Interpret: At frequencies well below resonance, displacement is in phase with the force (phi = 0). At resonance, phase lag is exactly pi/2. At high frequencies, displacement lags the force by pi radians.
Exam use: Tested via questions asking for the phase difference between driver and oscillator at resonance.
Draw: Experimental setup showing a tuning fork held above a resonance tube with adjustable water column height.
Interpret: Loud sound is heard when the air column length matches an odd multiple of a quarter wavelength, representing acoustic forced oscillation and resonance.
Exam use: Used in laboratory-based physics questions and practical theory exams to calculate velocity of sound.
Solved numericals
TGT core
A particle of mass 0.5 kg is subjected to a driving force F = 2 cos(10t) N. If the spring constant is 50 N/m and damping is negligible, find the steady-state amplitude of oscillation.
Calculate natural frequency omeg$a_0 = \sqrt(k / m) = \sqrt(50 / 0.5) = \sqrt(100) = 10 rad/s$.
Note that the driving frequency omega equals the natural frequency omega_0 (resonance condition).
Use the amplitude formula for zero damping: $A = (F_0 / m) / |\omega_0^2 - \omega^2|$. Wait, at exact resonance with zero damping, amplitude diverges. Let's provide a non-resonant frequency or include small damping to avoid infinity. Let's use driving frequency omeg$a = 8 rad/s instead$.
An oscillator has a natural frequency of 200 rad/s and a damping parameter gamm$a = 2 s^-1$. A harmonic force of amplitude 4 N and frequency 200 rad/s is applied. If the mass is 1 kg, calculate the steady-state amplitude at resonance.
The restoring force is strictly linear (obeys Hooke's law), which holds true only for small amplitudes of oscillation.
The external driving force is assumed to be purely sinusoidal with a constant frequency and amplitude.
The damping force is assumed to be directly proportional to the velocity (Stokes' damping).
Common misconceptions
Common error: The system oscillates at its natural frequency even after the transient state dies out.
Correct: In the steady state, the system oscillates exclusively at the driving frequency, not its natural frequency.
After transients decay, the external periodic force dictates the oscillation frequency entirely.
Common error: Maximum amplitude always occurs exactly when the driving frequency equals the natural frequency (omeg$a = \omega_0)$.
Correct: Maximum amplitude occurs at the resonant frequency omeg$a_r = \sqrt(\omega_0^2 - 2gamma^2)$, which is slightly lower than omega_0 in the presence of damping.
The damping term shifts the frequency at which the denominator of the amplitude expression is minimized.
Common error: At resonance, the phase lag between the driving force and displacement is zero.
Correct: At resonance, the phase lag phi is exactly pi/2 (90 degrees).
At resonance, the velocity is in phase with the driving force, meaning the displacement lags behind the force by pi/2.
Quick revision
Steady-state frequency equals the driving frequency omega.
Resonance amplitude is inversely proportional to damping gamma.
Quality factor $Q = \omega_0 / (2gamma) determines the sharpness of resonance$.
Phase lag at resonance is always pi/2 radians.
Must remember
The equation of motion for a forced damped oscillator is given by $\frac{d^2x}{dt^2} + 2\gamma\frac{dx}{dt} + \omega_0^2 x = \frac{F_0}{m} \cos(\omega t)$.
In the steady state, the system oscillates exclusively at the driving frequency $\omega$, with displacement $x(t) = A \cos(\omega t - \phi)$.
The phase lag $\phi$ is given by $\tan(\phi) = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}$. At resonance, $\phi = \pi/2$.
Amplitude resonance occurs at the resonant angular frequency $\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$, which is lower than the natural frequency $\omega_0$.
The Quality factor $Q$ measures the sharpness of resonance and is defined as $Q = \frac{\omega_0}{2\gamma} \approx \frac{\omega_0}{\Delta\omega}$.
Glossary
Forced OscillationsOscillations of a system under the influence of an external periodic driving force.
Damping Factor ($\gamma$)A parameter defined as $\frac{r}{2m}$ representing the rate of energy dissipation due to resistive forces.
Natural Frequency ($\omega_0$)The angular frequency $\sqrt{k/m}$ at which a system oscillates freely in the absence of damping and external forces.
Driving Frequency ($\omega$)The angular frequency of the external periodic force applied to the oscillator.
Phase Lag ($\phi$)The phase difference by which the displacement lags behind the external driving force.
Quality Factor ($Q$)A dimensionless parameter that characterizes the sharpness of resonance and energy storage relative to energy dissipation per cycle.
Natural vs Damped vs Forced Oscillations
Feature
Natural Oscillations
Damped Oscillations
Forced Oscillations
External Force
None
None
Periodic (e.g., $F_0 \cos(\omega t)$)
Amplitude
Constant
Decreases exponentially with time
Constant in steady state
Frequency
Natural frequency $\omega_0$
Modified frequency $\sqrt{\omega_0^2 - \gamma^2}$
Driving frequency $\omega$ in steady state
Energy
Conserved
Decreased due to friction/dissipation
Supplied continuously by external source
Practice Quiz
Choose an option to reveal the explanation.
1. What is the differential equation governing a forced damped harmonic oscillator?
Explanation: The general forced damped oscillator includes the restoring term $\omega_0^2 x$, the damping term $2\gamma \frac{dx}{dt}$, and the external driving force term $\frac{F_0}{m}\cos(\omega t)$.
2. In the steady state of a forced oscillation, what is the frequency of oscillation of the system?
Explanation: After the transient terms die out, the steady-state response of the system is entirely dictated by the external periodic force and hence oscillates at the driving frequency $\omega$.
3. What is the phase lag $\phi$ between the displacement and the driving force at resonance ($\omega = \omega_0$) for light damping?
Explanation: The phase angle is given by $\tan(\phi) = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}$. At resonance where $\omega = \omega_0$, the denominator becomes zero, making $\tan(\phi) = \infty$, which means $\phi = \pi/2$.
4. The resonant angular frequency $\omega_r$ for amplitude resonance in a damped system is given by:
Explanation: Amplitude resonance occurs when the denominator of the steady-state amplitude expression is minimized with respect to $\omega$, yielding $\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$.
5. What happens to the phase lag $\phi$ at very high driving frequencies ($\omega \gg \omega_0$)?
Explanation: When $\omega \gg \omega_0$, the denominator $\omega_0^2 - \omega^2$ becomes largely negative, leading $\tan(\phi)$ to approach negative zero from the negative side, meaning $\phi$ approaches $\pi$.
6. How is the Quality factor ($Q$) of a resonant system defined in terms of natural frequency and damping?
Explanation: The Quality factor $Q$ is defined as $Q = \frac{\omega_r}{2\gamma} \approx \frac{\omega_0}{2\gamma}$ for light damping.
7. Which parameter determines the bandwidth (full width at half maximum) of the power resonance curve?
Explanation: The full width at half maximum (bandwidth) of the power resonance curve is given by $\Delta \omega = 2\gamma$, meaning the damping parameter directly controls the sharpness and width of the resonance.
8. What is the approximate maximum amplitude at resonance ($A_{max}$) expressed using the Quality factor $Q$?
Explanation: At resonance with light damping, the peak amplitude is $A_{max} = \frac{Q F_0}{k}$, showing that a high Q-factor results in a drastically magnified amplitude.
9. At very low driving frequencies ($\omega \ll \omega_0$), what is the phase lag $\phi$?
Explanation: At low driving frequencies where $\omega \ll \omega_0$, the system's displacement is in phase with the driving force, so $\phi \approx 0$.
10. A high Quality factor ($Q$) implies which of the following characteristics?
Explanation: A high Q-factor corresponds to very light damping ($\gamma$ is small), which results in a sharp, narrow resonance peak and high amplitude magnification at resonance.
11. What is the physical meaning of the transient term in the solution of the forced oscillation equation?
Explanation: The transient part depends on initial conditions and decays exponentially with time due to the damping term $r \frac{dx}{dt}$, leaving only the steady-state response.
12. What are the dimensions or units of the damping factor $\gamma = \frac{r}{2m}$?
Explanation: The damping factor $\gamma$ has units of inverse time ($s^{-1}$), appearing in the exponent of damping terms like $e^{-\gamma t}$.
13. Under what condition does the amplitude resonance frequency $\omega_r$ become approximately equal to the natural frequency $\omega_0$?
Explanation: Since $\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$, for very light damping ($\gamma \approx 0$), the subtraction term is negligible, so $\omega_r \approx \omega_0$.
14. What does the average power absorbed by a forced oscillator depend upon in the steady state?
Explanation: The average power formula $\langle P \rangle = \frac{1}{2} \frac{F_0^2 \gamma \omega^2}{m [(\omega_0^2 - \omega^2)^2 + 4\gamma^2\omega^2]}$ clearly shows its dependence on $F_0$, $\gamma$, and $\omega$.
15. Which misconception about forced oscillations is correct regarding the frequency at resonance?
Explanation: Due to damping, the peak amplitude occurs at $\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$, which is slightly lower than the natural frequency $\omega_0$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. The steady-state amplitude of a forced oscillator with mass $m$, driving force $F_0$, natural frequency $\omega_0$, driving frequency $\omega$, and damping factor $\gamma$ is given by:
Explanation: This is the standard expression derived for the steady-state amplitude of a linear damped harmonic oscillator driven by a sinusoidal force.
2. The phase lag $\phi$ of displacement behind the driving force in a forced oscillator satisfies which relation?
Explanation: The phase relationship derived from the steady-state solution gives $\tan(\phi) = \frac{2\gamma\omega}{\omega_0^2 - \omega^2}$.
3. If the quality factor $Q$ of a resonating system is 50 and its natural angular frequency is $500\text{ rad/s}$, what is the damping factor $\gamma$?
Explanation: Using $Q = \frac{\omega_0}{2\gamma}$, we get $2\gamma = \frac{\omega_0}{Q} = \frac{500}{50} = 10$, which gives $\gamma = 5\text{ s}^{-1}$.
4. For a lightly damped forced oscillator, the resonant frequency $\omega_r$ is approximately equal to:
Explanation: When damping is very light ($\gamma \approx 0$), the expression $\omega_r = \sqrt{\omega_0^2 - 2\gamma^2}$ simplifies to $\omega_r \approx \omega_0$.
5. The bandwidth (full width at half maximum) of a power resonance curve is determined by which of the following?
Explanation: The frequency interval between the two half-power points is given by $\Delta \omega = \omega_2 - \omega_1 = 2\gamma$.
6. At resonance, what is the relation between velocity and the external driving force in a forced oscillator?
Explanation: Since displacement lags the force by $\pi/2$ at resonance ($\phi = \pi/2$), velocity (which is the time derivative of displacement) leads displacement by $\pi/2$, making velocity exactly in phase with the driving force.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which part of the general solution of a forced oscillation differential equation dies out as time progresses?
Explanation: The transient part depends on initial conditions and contains an exponential damping factor that causes it to die out, leaving only the steady-state response.
2. What is the phase lag $\phi$ at resonance when $\omega = \omega_0$?
Explanation: At resonance, $\omega = \omega_0$, making $\tan(\phi) = \infty$, which corresponds to a phase lag of $\phi = \pi/2$.
3. If the mass of a system is $1\text{ kg}$, spring constant is $100\text{ N/m}$, and damping coefficient $r = 2\text{ kg/s}$, what is the damping factor $\gamma$?
Explanation: The damping factor is $\gamma = \frac{r}{2m} = \frac{2}{2(1)} = 1\text{ s}^{-1}$.
4. The sharpness of resonance is directly measured by:
Explanation: The Quality factor $Q$ is defined specifically as a measure of the sharpness of resonance.
5. For a forced oscillator, when the driving frequency is much higher than the natural frequency ($\omega \gg \omega_0$), the steady-state amplitude is inversely proportional to:
Explanation: When $\omega \gg \omega_0$, the denominator in the amplitude expression is dominated by $\omega^2$ squared, making $A \approx \frac{F_0}{m \omega^2}$.
6. What is the expression for peak amplitude at resonance $A_{max}$ in terms of $F_0$, $m$, $\gamma$, and $\omega_r$?
Explanation: Substituting $\omega = \omega_r$ into the general amplitude formula gives $A_{max} = \frac{F_0}{2m\gamma\omega_r}$.
7. Which of the following statements about power absorption at resonance is correct?
Explanation: At resonance, the velocity is completely in phase with the driving force, resulting in maximum rate of work done and maximum average power absorption.
8. If a system has a natural frequency of $100\text{ rad/s}$ and a damping factor $\gamma = 2\text{ s}^{-1}$, what is its Q-factor?
Explanation: Using $Q = \frac{\omega_0}{2\gamma} = \frac{100}{2(2)} = \frac{100}{4} = 25$? Wait, $2\gamma = 4$, so $100/4 = 25$. Correct option is 25 (index 0).
9. In forced oscillations, what is the effect of decreasing damping (making $\gamma$ smaller)?
Explanation: Smaller damping decreases energy dissipation, leading to a higher Q-factor, which means a sharper, taller resonance peak.
10. Which of the following describes the correct unit and dimension of the Quality factor ($Q$)?
Explanation: The Quality factor $Q$ is a ratio of frequencies or energies, making it a dimensionless quantity.