Student Details







Forced Harmonic Oscillator — Obtaining the Solution

A forced harmonic oscillator is a system that oscillates under the influence of both a restoring force and an external periodic driving force. Typical examples include an electric circuit driven by an AC source, or a pendulum driven by a periodic torque.

1. Equation of Motion

Consider a mass \( m \) attached to a spring of stiffness \( k \), subjected to a damping force proportional to velocity, and driven by an external periodic force \( F_0 \cos(\Omega t) \).

The net force is given by: \[ F_{\text{net}} = -kx - c\dot{x} + F_0 \cos(\Omega t) \] Applying Newton’s second law: \[ m\ddot{x} + c\dot{x} + kx = F_0 \cos(\Omega t) \]

Dividing through by \( m \): \[ \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = f_0 \cos(\Omega t) \] where: \[ \omega_0 = \sqrt{\frac{k}{m}}, \quad 2\beta = \frac{c}{m}, \quad f_0 = \frac{F_0}{m} \]

2. General Solution

The total solution of this differential equation consists of two parts: \[ x(t) = x_h(t) + x_p(t) \] where:

3. Homogeneous Solution (Transient Part)

The homogeneous equation is: \[ \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = 0 \] which is the same as the damped oscillator equation.

Its solution depends on the damping condition: \[ x_h(t) = A e^{-\beta t} \cos(\omega_d t + \phi) \] where \( \omega_d = \sqrt{\omega_0^2 - \beta^2} \).

This term decays with time due to the exponential factor \( e^{-\beta t} \) and eventually vanishes, leaving only the steady-state solution.

4. Particular Solution (Steady-State Response)

To find the steady-state response, assume a trial solution of the form: \[ x_p(t) = X \cos(\Omega t - \delta) \] where:

Substitute this trial solution into the equation of motion:

\[ -\Omega^2 X \cos(\Omega t - \delta) + 2\beta\Omega X \sin(\Omega t - \delta) + \omega_0^2 X \cos(\Omega t - \delta) = f_0 \cos(\Omega t) \]

Using trigonometric identities and comparing coefficients of \( \cos(\Omega t) \) and \( \sin(\Omega t) \), we obtain two equations: \[ X(\omega_0^2 - \Omega^2) = f_0 \cos\delta \quad \text{and} \quad 2\beta\Omega X = f_0 \sin\delta \]

5. Amplitude of Steady-State Oscillation

Squaring and adding the two equations: \[ X^2 \big[ (\omega_0^2 - \Omega^2)^2 + (2\beta\Omega)^2 \big] = f_0^2 \] Hence, \[ \boxed{X = \frac{f_0}{\sqrt{(\omega_0^2 - \Omega^2)^2 + (2\beta\Omega)^2}}} \]

This shows that \( X \) depends on both the driving frequency \( \Omega \) and the damping constant \( \beta \).

6. Phase Difference

The phase lag \( \delta \) is obtained from: \[ \tan\delta = \frac{2\beta\Omega}{\omega_0^2 - \Omega^2} \]

7. Resonance

The amplitude \( X \) reaches a maximum when the denominator is minimum, i.e. when: \[ \frac{dX}{d\Omega} = 0 \] This occurs approximately at: \[ \boxed{\Omega_r = \sqrt{\omega_0^2 - 2\beta^2}} \] known as the resonance frequency.

For weak damping (\( \beta \ll \omega_0 \)), \( \Omega_r \approx \omega_0 \).

8. Velocity and Energy of Forced Oscillator

The velocity of oscillation is: \[ v(t) = -\Omega X \sin(\Omega t - \delta) \] and the average power absorbed from the driving force is: \[ \langle P \rangle = \frac{1}{2} F_0 \Omega X \sin\delta = \frac{F_0^2}{2m} \frac{2\beta\Omega^2} {(\omega_0^2 - \Omega^2)^2 + (2\beta\Omega)^2} \]

The power absorbed is maximum when \( \Omega = \omega_0 \), that is, at resonance.

9. Summary Table

Quantity Expression
Equation of motion \( \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = f_0 \cos(\Omega t) \)
Steady-state amplitude \( X = \dfrac{f_0}{\sqrt{(\omega_0^2 - \Omega^2)^2 + (2\beta\Omega)^2}} \)
Phase lag \( \tan\delta = \dfrac{2\beta\Omega}{\omega_0^2 - \Omega^2} \)
Resonance frequency \( \Omega_r = \sqrt{\omega_0^2 - 2\beta^2} \)
Average power absorbed \( \langle P \rangle = \dfrac{F_0^2}{2m} \dfrac{2\beta\Omega^2} {(\omega_0^2 - \Omega^2)^2 + (2\beta\Omega)^2} \)

10. Physical Interpretation

Few questions like question 5 is out of Syllabus.

Q1. The differential equation of a forced damped harmonic oscillator is:




Q2. In the equation \( \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = f_0 \cos(\Omega t) \), the term \( 2\beta \) represents:




Q3. The particular (steady-state) solution of a forced oscillator is of the form:




Q4. The amplitude of steady-state oscillation is given by:




Q5. The phase lag \( \delta \) between displacement and force is given by:




Q6. At resonance, the driving frequency \( \Omega \) is approximately equal to:




Q7. When \( \Omega = \omega_0 \), the phase difference between force and displacement is:




Q8. The transient part of the forced oscillator solution:




Q9. The steady-state part corresponds to:




Q10. The average power absorbed by the oscillator is:




Q11. Resonance in a mechanical system occurs when:




Q12. A bridge collapses when soldiers march in step because:




Q13. Which quantity determines how sharply resonance occurs?




Q14. The quality factor is defined as:




Q15. A high quality factor indicates:




Q16. When wind causes oscillations in tall buildings, the phenomenon is known as:




Q17. Earthquakes are dangerous for tall buildings when:




Q18. In resonance, which physical quantity reaches a maximum?




Q19. In a forced oscillator, when \( \Omega \gg \omega_0 \), the displacement is:




Q20. The bridge collapse in Tacoma Narrows (1940) was due to:




Q21. The amplitude–frequency curve of a forced oscillator becomes broader when:




Q22. At low driving frequencies, the displacement and the force are:




Q23. In the absence of damping, the amplitude at resonance is:




Q24. Increasing damping causes the resonance peak to:




Q25. The phase angle \( \delta \) changes from 0 to \( \pi \) as:




Q26. For small damping, the resonance frequency \( \Omega_r \) is approximately:




Q27. The oscillations produced by an alternating voltage across a resistor, inductor and capacitor are:




Q28. The amplitude of vibration of a building during an earthquake depends mainly on:




Q29. A lightly damped oscillator driven at resonance absorbs:




Q30. The main reason soldiers break step while crossing bridges is to:




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