OSCILLATIONS-QnA

Oscillatory motion is one of the central themes of Class 11 Physics and appears repeatedly in higher mechanics, waves, optics, and modern physics. The QnA set you are using systematically covers every exam-relevant concept of the NCERT chapter “Oscillations”, starting from basic definitions and building up to derivations, numericals, and descriptive discussions on real-life applications. The 1-mark and short-answer questions consolidate core ideas like period, frequency, phase, and graphical interpretation of SHM, ensuring that students can recall and apply formulas instantly in objective and very-short-answer sections of board papers and entrance exams. Medium and long answers deal with derivations of SHM equations, energy relations, time period of a simple pendulum, damped motion, forced oscillations, resonance, and beat phenomena so that learners can write stepwise, high-scoring solutions in descriptive questions.

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Maths

TRIGONOMETRIC FUNCTIONS-Exercise 3.1

Exercise • Jan 2026

Trigonometric Functions form a crucial foundation of higher mathematics and play a vital role in physics, engineering, astronomy, and real-life proble...

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Maths

TRIGONOMETRIC FUNCTIONS-True/False

True-False • Jan 2026

This structured collection of True and False questions on Trigonometric Functions is designed to support both two-session and four-session instruction...

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OSCILLATIONS

by Academia Aeternum

1-2 liner Questions

Q1: Define Simple Harmonic Motion (SHM)

A motion in which the restoring force is directly proportional to the displacement and acts in the opposite direction to the displacement. Mathematically, F = -kx where k is the spring constant.

Q2: What is the SI unit of angular frequency?

The SI unit of angular frequency \(?)\ is radian per second (rad/s) or simply s?¹

Q3: Define time period of oscillation

Time period is the time taken by an oscillating body to complete one full oscillation and return to the same position with the same velocity. It is denoted by T.

Q4: What is the relationship between frequency and time period?

Frequency is the reciprocal of time period. \(f = \dfrac{1}{T}\), where f is frequency in Hertz and \{T\} is time period in seconds.

Q5: Define amplitude in oscillatory motion

Amplitude is the maximum displacement of an oscillating body from its mean or equilibrium position. It represents the farthest distance the body travels from the equilibrium point.

Q6: What is the phase of an oscillator?

Phase is the stage or state of oscillation of a particle at a given instant. It determines the position and velocity of the particle at that particular time in its oscillatory cycle.

Q7: State the condition for two oscillators to be in resonance

Two oscillators are in resonance when their natural frequencies are equal or nearly equal, leading to maximum amplitude of oscillation.

Q8: What is damping in oscillatory motion?

Damping is the dissipation of energy from an oscillating system due to friction or other resistive forces, resulting in gradual decrease in amplitude of oscillation.

Q9: Write the equation of SHM in standard form

The standard equation of SHM is \(x(t) = A \sin(\omega t + \phi)\) or \(x(t) = A cos(\omega t + \phi)\), where \(A\) is amplitude, \(\omega \)is angular frequency, \(t\) is time, and \(\phi\) is initial phase.

Q10: What is the angular frequency of a spring-mass system?

The angular frequency of a spring-mass system is \(\omega = sqrt{\dfrac(k}{m}\), where \(k\) is spring constant and \(m\) is mass of the object.

Q11: Define restoring force

Restoring force is the force that always acts to return an oscillating body to its equilibrium position. It is proportional to displacement and opposite in direction.

Q12: What is the velocity of a particle in SHM at the equilibrium position?

The velocity is maximum at the equilibrium position. Using \(v = ±\omega \sqrt{(A^2 - x^2)}\), when \(x = 0,\, v_max = \omega A\)

Q13: What is the acceleration of a particle in SHM at the extreme positions?

The acceleration is maximum at the extreme positions. When displacement \(x = ±A\), acceleration \(a_max = ±\omega^2 A\) (maximum magnitude)

Q14: What is the relation between time period and length of a simple pendulum?

The time period of a simple pendulum is \(T = 2p\sqrt{(\dfrac{L}{g})}\), where L is length of pendulum and g is acceleration due to gravity.

Q15: Define forced oscillation

Forced oscillation occurs when an external periodic force is applied to an oscillating system. The frequency of oscillation equals the frequency of the applied external force.

Short answer Questions

Q1: A particle executing SHM has displacement x = 4sin(pt + p/4) cm. Find its amplitude, time period, and frequency.

Comparing with \[x = A\, \sin(\omega t + \phi)\] Amplitude A = 4 cm, Angular frequency \(\omega = p\ rad/s\),
Time period \[\begin{aligned} T& = \dfrac{2p}{p} &\\\\&= 2p/p \\\\&= 2 \text{ seconds} \end{aligned}\] Frequency \[f = \dfrac{1}{T} = 0.5\ \text{Hz}\]

Q2: Explain why the motion of a particle in a circular path can be considered as a combination of two SHMs.

When a particle moves in a circular path with constant angular velocity, its x-component and y-component of displacement separately execute SHM with the same amplitude and frequency but different phases. The projection of circular motion on any diameter gives SHM.

Q3: A mass attached to a spring performs SHM. If the mass is increased, how does it affect the time period?

Time period \(T = 2p\sqrt{(\dfrac{m}{k})}\). Since \(T\) is directly proportional to \(sqrt{m}\), increasing mass increases the time period. For example, if mass is quadrupled, time period doubles.

Q4: Compare the energy at equilibrium position and extreme position in SHM.

At equilibrium position: Kinetic energy is maximum \[KE = \dfrac{1}{2}m\omega^2A^2\] Potential energy is zero, Total energy is maximum kinetic. At extreme position: Kinetic energy is zero, Potential energy is maximum \[PE = \dfrac{1}{2}m\omega^2A^2\] Total energy equals maximum potential energy. Total energy remains constant.

Q5: What is the difference between oscillation and vibration?

Oscillation is a general periodic motion, while vibration refers to rapid oscillations, typically at high frequencies. Vibration usually implies faster motion with smaller amplitudes and higher frequencies compared to oscillation.

Q6: A simple pendulum of length L is taken from Earth to the Moon. How does its time period change?

Time period \[T = 2p\sqrt{\dfrac{L}{g}}\] Since \[g_{moon} \approx \dfrac{g_{earth}}{6}\] the time period on Moon will be 6 times greater than on Earth. This means the pendulum oscillates more slowly on the Moon.

Q7: Write the equation for velocity and acceleration in SHM if position is x = A sin(?t)

For \[x = A sin(\omega t)\] Velocity \[v = \dfrac{dx}{dt} = A\omega \cos(\omega t)\] Acceleration \[\begin{aligned}a &= \dfrac{dv}{dt}\\\\&= -A\omega^2 \sin(\omega t) \\\\&= -\omega^2 x\end{aligned}\] The acceleration is always opposite to displacement and proportional to it.

Q8: Explain the concept of resonance and its practical applications.

Resonance occurs when external driving frequency equals the natural frequency of a system, causing maximum amplitude. Applications: Radio tuning circuits, MRI machines, architectural design for preventing bridge collapse, microwave ovens, and musical instruments. Careful handling needed to prevent damage from excessive vibrations.

Q9: How does the velocity of a particle change as it moves from equilibrium to extreme position?

As the particle moves from equilibrium position to extreme position, velocity decreases linearly with displacement. At equilibrium, \(v\) is maximum \(v = \omega A\). At extreme position, \(v\) becomes zero. This relationship is given by \[v = \omega\sqrt{A^2 - x^2}\]

Q10: A particle performs SHM with amplitude 2 cm and frequency 3 Hz. Calculate its maximum speed.

Given: A = 2 cm = 0.02 m,
f = 3 Hz.
Angular frequency \[\begin{aligned}\mathrm{\omega} &= 2\pi f \\&= 2\pi (3) \\&= 6\pi\ rad/s\end{aligned}\]
Maximum speed \[\begin{aligned} v_{max} &= \omega A \\&= 6\pi × 0.02 \\&= 0.12\pi \\& \approx 0.377\ m/s \end{aligned}\]

Long answer Questions

Q1: Derive the equation for the time period of a simple pendulum. Explain the factors it depends upon.

Derivation:
For a simple pendulum, restoring force is proportional to angle: \[\begin{aligned}\tau &= -mg L \sin \theta \\\\&\approx -mg L \theta\end{aligned}\] for small angles.
Angular acceleration \(a = -g/L × \theta\).
Comparing with \(a = -\omega^2 \theta\), we get \[\omega^2 =g/L\]
so \[\omega = \sqrt{\dfrac{g}{L}}\] Time period \[\begin{aligned}T &= \dfrac{2\pi}{\omega} \\\\&= 2p\sqrt{\dfrac{L}{g}}\end{aligned}\] Factors: \(T\) depends directly on \(\sqrt{L}\) and inversely on \(\sqrt{g}\). T is independent of mass and amplitude (for small angles). This is the isochronous property of pendulum.

Q2: Explain the concept of phase difference. How does it affect the resultant motion when two SHMs of same frequency are superposed?

Phase difference \((\Delta \phi)\) is the difference between the phases of two oscillating particles at any given time.

When two SHMs of same amplitude A and frequency \(\mathrm{\omega}\) with phase difference \(\Delta \phi\) are superposed: \[x_1 = A \sin(\omega t)\] and \[x_2 = A \sin(\omega t + \Delta \phi)\] resultant amplitude \[R = 2A \cos(\Delta \phi/2)\]

Special cases:
(1) \((\Delta \phi)\) = 0° (in phase): R = 2A (constructive interference),
(2) \((\Delta \phi)\)= 180° (out of phase): R = 0 (destructive interference),
(3) \((\Delta \phi) = 90^\circ: R = A\,\sqrt{v}\) (partial reinforcement). The resultant phase is \((\omega t + \Delta \phi/2)\).

Q3: A body of mass 1 kg is attached to a spring of spring constant 100 N/m. If the body is displaced by 10 cm from equilibrium and released, find: (i) Time period (ii) Maximum velocity (iii) Maximum acceleration

Time period:
\[\begin{aligned}T &= 2\pi\sqrt{\dfrac{m}{k}} \\\\&= 2\pi\sqrt{\dfrac{1}{100}} \\\\&= \dfrac{2\pi}{10} \\\\&= \dfrac{\pi}{5} \\\\&\approx 0.628\,s\end{aligned}\]
Amplitude
A = 10 cm = 0.1 m.
Angular frequency
\[\begin{aligned}\omega &= \sqrt{\dfrac{k}{m}} \\\\&= \sqrt{\dfrac{100}{1}} \\\\&= 10\ rad/s\end{aligned}\]
Maximum velocity
\[\begin{aligned}v_{max} &= \omega A\\&= 10 × 0.1 \\&= 1 m/s\end{aligned}\]
Maximum acceleration
\[\begin{aligned}a_{max} &= \omega^2 A \\&= 100 × 0.1 \\&= 10 ms_{-2}\end{aligned}\] These values remain constant throughout the motion as the system has no damping.

Q4: What is the difference between free oscillation and forced oscillation? Explain resonance in context of forced oscillation.

Free oscillation: System oscillates with its natural frequency after initial disturbance, without external force. Amplitude remains constant (ideal case). Forced oscillation: External periodic force continuously drives the system. The frequency of oscillation equals the driving frequency, not the natural frequency. Resonance: Occurs when driving frequency equals natural frequency of the system. At resonance: Amplitude becomes maximum, energy transfer is most efficient, phase difference between driving force and displacement becomes 90°. Real-world examples include shattering of glass by sound waves, collapse of bridges due to wind, and tuning of radio circuits. Quality factor Q determines sharpness of resonance peak.

Q5: Establish the relationship between kinetic energy, potential energy, and total energy in SHM. Show that total energy remains constant.

For SHM:
\[x = A \sin(\omega t)\] \[v = A\omega \cos(\omega t)\]
Kinetic energy:
\[\begin{aligned}KE &=\dfrac{1}{2}mv^2 \\\\&= \dfrac{1}{2}m(A\omega)^2 \cos^2(\omega t) \\\\&= \dfrac{1}{2}m\omega^2 A^2 \cos^2(\omega t)\end{aligned}\]
Potential energy:
\[\begin{aligned}PE &= \dfrac{1}{2}kx^2 \\\\&= \dfrac{1}{2}m\omega^2A^2 \sin^2(\omega t)\\ \text{[since } k &= m\omega^2]\end{aligned}\]
Total energy:
\[\begin{aligned}E &= KE + PE \\\\&= \dfrac{1}{2}m\omega^2A^2[cos^2(\omega t) + sin^2(\omega t)] \\\\&= \dfrac{1}{2}m\omega^2A^2 = \text{constant}\end{aligned}\] At equilibrium (x = 0):
KE is maximum, PE = 0.
At extreme positions (x = ±A): PE is maximum, KE = 0.
At intermediate positions: both KE and PE are non-zero but their sum remains constant, demonstrating conservation of energy in SHM.

Descriptive Questions

Q1: Derive the equation for acceleration in terms of displacement for a particle executing SHM. Starting from Newton's second law, show that the motion satisfies the differential equation of SHM and obtain the general solution. Explain the physical significance of each parameter in the solution.

Derivation from Newton's Second Law: For SHM, the restoring force is \[F = -kx\] By Newton's second law, \[F = ma\] therefore: \[ma = -kx\] which gives \[a = -\dfrac{k}{m}x\] Let \[\omega^2 = \dfrac{k}{m}\] then \[a = -\omega^2x\] Since acceleration \[a = \dfrac{d^2x}{dt^2}\] we have: \[\dfrac{d^2x}{dt^2} = -\omega^2x,\] or \[\dfrac{d^2x}{dt^2} + \omega^2 x = 0\] This is the differential equation of SHM.

Solution:
The general solution to this differential equation is \[x(t) = A \sin(\omega t + \phi)\] or \[x(t) = A \cos(\omega t + \phi)\] where \(A\) and \(\phi\) are constants determined by initial conditions.
Physical Significance:
(1) Amplitude A: Maximum displacement from equilibrium; represents the energy stored in the system.
(2) Angular frequency \(\omega = 2\pi f\): Determines how fast the oscillation occurs; \(\omega = \sqrt{\dfrac{k}{m}}\) for spring-mass system.
(3) Phase constant \(\phi\): Determines initial position and velocity at \(t = 0\); \(\phi\) determines time reference.
(4) The negative sign in the differential equation indicates that acceleration is always opposite to displacement, ensuring restoring force.
(5) The quadratic relation between acceleration and displacement \((a \propto -x)\) is fundamental to SHM and distinguishes it from other types of periodic motion.
Velocity and Acceleration Relations:
From \[\begin{aligned} x = A \sin(\omega t + \phi)\end{aligned}\] we get: \[\begin{aligned} v &= \dfrac{dx}{dt} \\\\&= A\omega \cos(\omega t + \phi)\end{aligned}\] and \[\begin{aligned} a &= \dfrac{dv}{dt} \\\\&= -A\omega^2 \sin(\omega t + \phi) \\\\&= -\omega^2x \end{aligned} \] These relations show that velocity leads displacement by 90° in phase, and acceleration leads velocity by 90° (or leads displacement by 180°). The energy in the system oscillates between kinetic and potential forms but remains constant overall.

Q2: Explain the concept of damped oscillations. Derive the differential equation for damped SHM and discuss different types of damping. How does damping affect amplitude, frequency, and energy of oscillation? What are the practical implications?

Damped Oscillations:
Damped oscillations occur when energy is continuously dissipated from an oscillating system due to friction, air resistance, or other resistive forces. The amplitude decreases exponentially with time.
Differential Equation:
The restoring force is \(-kx\) and damping force is \(-b(\frac{dx}{dt})\), where b is damping coefficient.
By Newton's second law: \[ \begin{aligned} m\left(\dfrac{d^2x}{dt^2}\right) = -kx - b\left(\dfrac{dx}{dt}\right) \end{aligned} \] Rearranging: \[ m\left(\dfrac{d^2x}{dt^2}\right) + b\left(\dfrac{dx}{dt}\right) + kx = 0 \] or \[\dfrac{d^2x}{dt^2} + (b/m)\left(\dfrac{dx}{dt}\right) + \left(\dfrac{k}{m}\right)x = 0\]

Let \(\dfrac{b}{m}\) and \(\omega_0^2 = \dfrac{k}{m}\), then:

\[\dfrac{d^2x}{dt^2} + 2\gamma \left(\dfrac{dx}{dt}\right) + \omega_0^2x = 0\]

This is the differential equation of damped SHM.

Types of Damping:
(1) Underdamped \(\gamma < \omega0\): System oscillates with decreasing amplitude.
Solution:
\[ A(t) = A_0e^{(-2\gamma t)} \sin(\omega't + \phi)\] where
\[\begin{aligned} \omega'=\sqrt{(\omega_0^2 - \gamma^2)} < \omega_0\end{aligned}\]
(2) Critically damped \((\gamma=\omega_0)\): System returns to equilibrium without oscillating in minimum time. Used in shock absorbers.
(3) Overdamped \((\gamma> \omega_0)\): System returns very slowly without oscillating.

Solution involves exponential terms only.

Effect on Oscillation Parameters:
(1) Amplitude decreases as \(A(t) = A_0e^{(-2\gamma t)}\),
(2) Frequency decreases: \(\omega' = \sqrt{(\omega_0^2 - \gamma^2)} < \omega_0\),
(3) Energy decreases exponentially: \(E_{(t)}=E_0e^{(-2\gamma t)}\),
(4) Time period increases gradually.
Energy Dissipation:

In each complete oscillation, energy is lost to the environment. The rate of energy loss is: dE/dt=-2?mvA(t) sin²(?t). The quality factor Q=?0/2? measures the ratio of energy stored to energy dissipated per cycle.

Practical Implications:
(1) Car shock absorbers use critical damping for comfort and control.
(2) Seismic detectors require appropriate damping to record ground motion accurately.
(3) Watches and clocks need minimal damping to maintain accuracy.
(4) Building design incorporates damping to prevent resonance during earthquakes.
(5) In musical instruments, slight damping gives natural decay of sound.

Understanding damping is crucial for designing systems that must either maintain or eliminate oscillations depending on application."

Q3: Discuss the phenomenon of resonance in forced oscillations. Derive the expression for amplitude at resonance. What factors determine the sharpness of resonance peak? Explain how resonance can be beneficial and harmful in real-world applications.

Forced Oscillations and Resonance:
When an external periodic force \(F(t) = F_0 \cos(\omega t)\) is applied to a damped oscillator, the equation of motion becomes: \[m\left(\dfrac{d^2x}{dt^2}\right) + b\left(\dfrac{dx}{dt}\right) + kx = F_0 \cos(\omega t)\] After transient effects die out, the system oscillates at the driving frequency ? (not the natural frequency ?0). The steady-state solution is \[x(t) = A \cos(\omega t - \delta)\] where \(A\) is amplitude and \(delta) is phase lag.
Amplitude Expression:
\[A(\omega) = \dfrac{\left(\dfrac{F_0}{m}\right)}{\sqrt{[(\omega_0^2 - \omega^2)² + (2\gamma \omega)^2]}}\] where \(\gamma = \dfrac{b}{2m}\), \(\omega_0 = \sqrt{\dfrac{k}{m}}\).
At resonance \[\omega = \omega_0: A_{max} = \dfrac{F_0}{2m\gamma\omega_0} = \dfrac{F_0}{b\omega_0}\] The amplitude depends inversely on damping coefficient b.
Sharpness of Resonance:
The quality factor Q = ?0/(2?) = m?0/b determines sharpness. (1) High Q (low damping): Sharp resonance peak, large amplitude at resonance, narrow frequency range. (2) Low Q (high damping): Broad resonance peak, lower amplitude, wide frequency range. Bandwidth ?? = ?0/Q. For high Q systems, resonance occurs at ? ˜ ?0. Phase lag at resonance is 90°.
Phase Relationship:
At resonance, velocity is in phase with driving force (maximum energy transfer), while displacement lags force by 90°. This explains why energy absorption is maximum at resonance.
Beneficial Applications:
(1) Radio tuning: Receiver resonates at desired frequency to select specific stations.
(2) Musical instruments: Resonance chambers amplify specific frequencies for rich sound.
(3) Wireless power transfer: Systems designed for resonance enable efficient energy transmission.
(4) Medical imaging: MRI uses nuclear resonance to create detailed images. (5) Energy harvesting: Resonant structures efficiently convert vibrations to electrical energy.
Harmful Applications:
(1) Bridge collapse: Wind-induced resonance caused Tacoma Narrows Bridge collapse in 1940.
(2) Building damage: Earthquakes at natural frequency cause severe structural damage. (3) Machine vibration: Resonance in rotating machinery can cause bearing failure and noise.
(4) Electrical circuits: Resonance can cause voltage spikes and component failure.
(5) Acoustic feedback: Microphone and speaker resonance causes annoying feedback in public address systems.
(6) Resonance catastrophe: If damping is insufficient and driving force is large, amplitude can grow unbounded,
destroying the system.
Preventive Measures: Use damping to control resonance peak, design structures with natural frequencies away from expected driving frequencies, use vibration isolators, and incorporate frequency-selective filters.
Understanding resonance is essential for engineers and scientists designing systems that interact with periodic forces.

Q4: Analyze the superposition of two Simple Harmonic Motions with the same amplitude but slightly different frequencies. Derive the expression for the resultant motion and explain the phenomenon of beats. How can beats be applied in tuning musical instruments?

Consider two simple harmonic motions (SHMs) with the same amplitude \(A\) but slightly different angular frequencies \(\omega_1\) and \(\omega_2\) \((\omega_2 \approx \omega_1)\).

The individual displacements are:

\[ x_1(t) = A \sin(\omega_1 t), \quad x_2(t) = A \sin(\omega_2 t) \]

The resultant displacement is the sum:

\[ x(t) = x_1(t) + x_2(t) = A[\sin(\omega_1 t) + \sin(\omega_2 t)] \]

Use of Trigonometric Identity

Using the identity \(\sin P + \sin Q = 2 \sin\left(\dfrac{P+Q}{2}\right) \cos\left(\dfrac{P-Q}{2}\right)\), the resultant becomes

\[ x(t) = 2A \sin\left(\frac{\omega_1 + \omega_2}{2} t\right) \cos\left(\frac{\omega_1 - \omega_2}{2} t\right) \]

This can be written as

\[ x(t) = \Big[2A \cos\left(\frac{\omega_1 - \omega_2}{2} t\right)\Big] \sin\left(\frac{\omega_1 + \omega_2}{2} t\right) \]

Physical Interpretation

The motion can be viewed as an SHM of average angular frequency \[ \omega_{\text{avg}} = \frac{\omega_1 + \omega_2}{2} \] whose amplitude varies with time as \[ A_{\text{mod}}(t) = 2A \left| \cos\left(\frac{\omega_1 - \omega_2}{2} t\right)\right| \]

The cosine factor represents a slowly varying envelope that modulates the fast oscillation of frequency \(\omega_{\text{avg}}\).

Beats Phenomenon

Since \(\omega_1 \approx \omega_2\), the inequality \[ \frac{\omega_1 - \omega_2}{2} \ll \frac{\omega_1 + \omega_2}{2} \] holds, so the sine term represents rapid oscillations, while the cosine term changes slowly and produces beats.

The amplitude varies between \(0\) and \(2A\) with beat angular frequency \[ \omega_{\text{beat}} = \omega_1 - \omega_2 \] and the beat frequency (number of intensity maxima per second) is \[ f_{\text{beat}} = |f_1 - f_2| \]

Beat Period

The beat period is \[ T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{1}{|f_1 - f_2|} \]

During each beat period the amplitude goes from maximum to zero and back to maximum twice, producing two intensity maxima.

Conditions for Clear Beats
  • Frequencies should be close: \(\omega_1 \approx \omega_2\).
  • Amplitudes should be approximately equal.
  • The frequency difference should be small enough (typically \(0.5\)–\(3\) Hz) so that individual beats are clearly heard in music.
Application in Musical Instrument Tuning
  1. Two instruments are sounded simultaneously; if their frequencies differ, beats are heard.
  2. The beat frequency (beats per second) gives the magnitude of the frequency difference.
  3. When the beat frequency approaches zero, the instruments are in tune.
  4. Musicians adjust pitch by changing string tension (string instruments) or air-column length (wind instruments) until beats disappear.
  5. In orchestras, players compare their notes with a reference tuning fork (usually \(A = 440\) Hz) and adjust to remove beats.
  6. The ear can detect very small differences (as low as about \(0.5\) Hz) by counting beats, making this method more sensitive than direct comparison.
Mathematical Precision in Tuning

If the initial frequency difference is \(\Delta f\) and the musician changes the frequency by \(\delta f\), the new beat frequency is \[ f_{\text{beat,new}} = |\Delta f - \delta f| \]

Fine tuning corresponds to making \(f_{\text{beat,new}}\) very small; this works because phase variations (beats) are easier to detect than small absolute frequency changes.

Extensions to Other Domains

The beat principle is used in several areas:

  • Heterodyne receivers for frequency mixing in radio engineering.
  • Optical interferometry for precise frequency and wavelength measurements.
  • Acoustic engineering for noise cancellation and sound analysis.
  • Frequency synthesis and sound design in electronic music.

Q5: Derive the expression for total energy in Simple Harmonic Motion. Prove that total mechanical energy remains constant and is independent of the position of the particle. Show how energy transforms between kinetic and potential forms during oscillation. What is the average kinetic and potential energy over one complete cycle?

"

Displacement and Velocity

Consider a particle in SHM with displacement and velocity

\[ x(t) = A \sin(\omega t), \quad v(t) = \frac{dx}{dt} = A \omega \cos(\omega t), \] where \(A\) is the amplitude and \(\omega\) is the angular frequency.

Kinetic Energy

The instantaneous kinetic energy is

\[ KE(t) = \frac{1}{2} m v^2 = \frac{1}{2} m \left[A \omega \cos(\omega t)\right]^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t). \]

Potential Energy

For SHM, the restoring force is \(F = -kx\) with \(k = m \omega^2\). Taking potential energy zero at equilibrium, the instantaneous potential energy is

\[ PE(t) = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t). \]

Total Mechanical Energy

The total mechanical energy is the sum of kinetic and potential energies:

\[ E_{\text{total}} = KE + PE = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) + \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t). \]

Using \(\cos^2(\omega t) + \sin^2(\omega t) = 1\),

\[ E_{\text{total}} = \frac{1}{2} m \omega^2 A^2 = \text{constant}. \]

Since \(\omega = \sqrt{\dfrac{k}{m}}\), this can also be written as

\[ E_{\text{total}} = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 = \frac{1}{2} m (2 \pi f)^2 A^2. \]

Proof of Constancy

The expression \(\dfrac{1}{2} m \omega^2 A^2\) depends only on the mass \(m\), angular frequency \(\omega\), and amplitude \(A\), all of which remain fixed for a given oscillator.

Hence, the total mechanical energy remains constant during the motion and is independent of the instantaneous position of the particle.

Energy Transformation During Oscillation
  1. Equilibrium position \((x = 0)\): \(\cos(\omega t) = \pm 1,\ \sin(\omega t) = 0\). \[ KE_{\max} = \frac{1}{2} m \omega^2 A^2,\quad PE_{\min} = 0. \] All the energy is kinetic.
  2. Extreme positions \((x = \pm A)\): \(\cos(\omega t) = 0,\ \sin(\omega t) = \pm 1\). \[ KE_{\min} = 0,\quad PE_{\max} = \frac{1}{2} m \omega^2 A^2. \] All the energy is potential.
  3. Intermediate positions \((0 < |x| < A)\): Both \(KE\) and \(PE\) are non-zero, and energy keeps converting between kinetic and potential forms while the total remains constant.
Energy–Position Relationship

Using \(x\) as the instantaneous displacement, the potential energy is \[ PE(x) = \frac{1}{2} k x^2 = \frac{1}{2} m \omega^2 x^2, \] and total energy is \(E_{\text{total}} = \dfrac{1}{2} m \omega^2 A^2\).

Therefore, the kinetic energy as a function of position is \[ KE(x) = E_{\text{total}} - PE(x) = \frac{1}{2} m \omega^2 A^2 - \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (A^2 - x^2). \]

The corresponding speed is \[ v = \sqrt{\frac{2\,KE(x)}{m}} = \omega \sqrt{A^2 - x^2}. \]

Average Energy Over One Complete Cycle

The average kinetic energy over one period \(T\) is \[ \langle KE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t)\, dt. \] Since \(\langle \cos^2(\omega t) \rangle = \dfrac{1}{2}\),

\[ \langle KE \rangle = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{2} = \frac{1}{4} m \omega^2 A^2. \]

Similarly, the average potential energy is \[ \langle PE \rangle = \frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin^2(\omega t)\, dt = \frac{1}{4} m \omega^2 A^2. \]

Thus, \[ \langle KE \rangle + \langle PE \rangle = \frac{1}{4} m \omega^2 A^2 + \frac{1}{4} m \omega^2 A^2 = \frac{1}{2} m \omega^2 A^2 = E_{\text{total}}. \]

On average, the total energy is equally shared between kinetic and potential forms.

Physical Significance

This equal sharing (equipartition) of energy between kinetic and potential forms is a fundamental property of harmonic oscillators.

It plays an important role in quantum mechanics and statistical mechanics, helping to explain the heat capacity of solids (Dulong–Petit law) and the behavior of quantum oscillators.

Graphical Representation

When kinetic and potential energies are plotted against time, both vary sinusoidally with twice the frequency of displacement: \(KE \propto \cos^2(\omega t)\) and \(PE \propto \sin^2(\omega t)\).

The two curves are mirror images about the horizontal line at \(E_{\text{total}}/2\), illustrating continuous exchange of energy while the total remains constant.

Energy Conservation Implication

The constant total mechanical energy \(\dfrac{1}{2} m \omega^2 A^2\) equals the work done by an external agent in initially displacing the particle to amplitude \(A\).

This initial energy fully accounts for all subsequent motion, demonstrating the principle of conservation of mechanical energy in an ideal simple harmonic oscillator.

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Frequently Asked Questions

Oscillatory motion is the motion in which a body moves repeatedly to and fro about a fixed mean position under the action of a restoring force.

Periodic motion is a type of motion that repeats itself after equal intervals of time, called the time period.

All oscillatory motions are periodic, but not all periodic motions are oscillatory because oscillatory motion must occur about a mean position.

SHM is a special type of oscillatory motion in which the restoring force is directly proportional to the displacement from the mean position and acts towards it.

A motion is SHM if the restoring force or acceleration is proportional to displacement and opposite in direction, i.e., \(a \propto -x\).

The mean position is the equilibrium position about which a body oscillates and where the net force acting on it is zero.

Amplitude is the maximum displacement of the oscillating body from its mean position.

Time period is the time taken by a body to complete one full oscillation.

Frequency is the number of oscillations completed per second and is the reciprocal of the time period.

Angular frequency \(\omega\) is defined as \(\omega = 2\pi f\), where \(f\) is the frequency of oscillation.

Phase represents the state of oscillation of a particle at a given instant, determined by the argument of the sine or cosine function.

Phase difference is the difference in phase angles of two oscillatory motions at the same instant.

The general equation of SHM is \(x = A\cos(\omega t + \phi)\), where \(A\) is amplitude and \(\phi\) is phase constant.

Restoring force is the force that always acts towards the mean position and tends to bring the body back to equilibrium.

SHM is called harmonic because its displacement varies sinusoidally with time.

Velocity in SHM is given by \(v = \omega\sqrt{A^2 - x^2}\).

Velocity is maximum at the mean position.

Velocity is zero at the extreme positions.

Acceleration in SHM is given by \(a = -\omega^2 x)\.

Acceleration is maximum at the extreme positions.

Acceleration is zero at the mean position.

Total mechanical energy in SHM is constant and equal to \(\frac{1}{2}kA^2\).

Energy continuously transforms between kinetic and potential forms while total energy remains constant.

Kinetic energy is maximum at the mean position.

Potential energy is maximum at the extreme positions.

It is a mechanical system where a mass attached to a spring executes SHM when displaced from equilibrium.

Hooke’s law states that the restoring force of a spring is proportional to its extension or compression.

The time period is \(T = 2\pi\sqrt{\frac{m}{k}}\).

A simple pendulum consists of a point mass suspended by a light, inextensible string from a fixed support.

A pendulum executes SHM only for small angular displacements.

The time period is \(T = 2\pi\sqrt{\frac{l}{g}}\).

A seconds pendulum has a time period of 2 seconds.

No, the time period of a simple pendulum is independent of the mass of the bob.

For ideal SHM, the time period is independent of amplitude.

It is the distance between the point of suspension and the centre of mass of the bob.

A motion that repeats after equal intervals of time but does not satisfy the SHM condition is periodic but non-SHM.

Motion described by \(x = \sin^2 \omega t\) is periodic but not SHM.

Because it assumes no friction, no energy loss, and perfectly linear restoring forces.

Free oscillations occur when a system oscillates with its natural frequency without external forces.

Damped oscillations are oscillations in which amplitude decreases due to energy loss.

Natural frequency is the frequency with which a system oscillates when disturbed and left free.

SHM can be considered as the projection of uniform circular motion on a diameter.

Pendulum clocks, quartz watches, tuning forks, and spring balances use oscillation principles.

It forms the foundation for waves, sound, AC circuits, and many competitive exam problems.

Numerical problems, derivations, conceptual reasoning, assertion–reason, and graph-based questions.

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