Chapter 13: Oscillations Long Questions | physics Class 11 CBSE

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Chapter 13: Oscillations Long Questions | physics Class 11 CBSE | ByteNotes.in

Long Answer

Hard difficulty • Structured explanation

Hard

Question 1

Long Form

Derive the expressions for velocity and acceleration of a particle in SHM using both the reference circle method and the differentiation method. Compare the results.

Reference circle method for velocity: if P moves on a circle of radius A with speed ωA, its projection P′ on the x-axis has velocity v(t) = −ωA sin(ωt + φ); the negative sign indicates direction opposite to the positive x-axis.
Differentiation method for velocity: differentiating x(t) = A cos(ωt + φ) with respect to time gives v(t) = dx/dt = −ωA sin(ωt + φ), identical to the reference circle result.
Reference circle method for acceleration: the centripetal acceleration of P has magnitude ω²A directed toward centre O; its x-projection gives a(t) = −ω²A cos(ωt + φ) = −ω²x(t).
Differentiation method for acceleration: differentiating v(t) = −ωA sin(ωt + φ) gives a(t) = −ω²A cos(ωt + φ), confirming the reference circle result.
Key property from a(t) = −ω²x(t): acceleration is always proportional to displacement and directed opposite to it; at x = 0 acceleration is zero; at x = ±A it is maximum (ω²A).
Both methods yield identical results, confirming the consistency of the reference circle model with calculus-based analysis of SHM.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 2

Long Form

Derive an expression for the total mechanical energy of a particle executing SHM. Show that it is constant and analyse how energy is distributed between kinetic and potential forms.

Kinetic energy: K = ½mv² = ½m(ωA)² sin²(ωt + φ) = ½kA² sin²(ωt + φ), where k = mω²; this is zero at extreme positions and maximum (½kA²) at the mean position.
Potential energy: the spring force F = −kx is conservative; the associated potential energy is U = ½kx² = ½kA² cos²(ωt + φ); this is zero at mean position and maximum (½kA²) at extreme positions.
Total energy: E = K + U = ½kA²[sin²(ωt + φ) + cos²(ωt + φ)] = ½kA², using the trigonometric identity sin²θ + cos²θ = 1.
The total energy ½kA² is constant and independent of time, as expected for a conservative force, confirming energy conservation.
At x = 0 (mean position), all energy is kinetic (K = ½kA², U = 0); at x = ±A (extremes), all energy is potential (U = ½kA², K = 0).
At intermediate positions, energy is shared between K and U, with one increasing at the expense of the other, while their sum remains constant at ½kA².

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 3

Long Form

Prove that the motion of a simple pendulum is simple harmonic for small displacements and derive the formula for its time period.

A simple pendulum consists of a bob of mass m on a massless inextensible string of length L; when displaced by angle θ from vertical, the restoring torque about the support is τ = −mgL sin θ (negative sign because it opposes displacement).
By Newton's law for rotation: τ = Iα, so Iα = −mgL sin θ, giving α = −(mgL/I) sin θ.
For small θ (in radians), sin θ ≈ θ, so α ≈ −(mgL/I)θ; this is mathematically identical to the SHM equation a = −ω²x, confirming simple harmonic motion.
Comparing: ω² = mgL/I; since the string is massless, I = mL², giving ω = √(g/L).
Time period: T = 2π/ω = 2π√(L/g); this is independent of the mass m of the bob and independent of amplitude (for small angles).
For a seconds pendulum (T = 2 s) under g = 9.8 m/s², the length L = gT²/4π² = 9.8 × 4/4π² ≈ 1 m.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 4

Long Form

Explain how SHM can be viewed as the projection of uniform circular motion on a diameter. Describe what changes when the projection is taken on the y-axis instead of the x-axis.

A particle P moves uniformly on a circle of radius A (the reference circle) with angular speed ω; its projection P′ on the x-axis has position x(t) = A cos(ωt + φ), which is the defining equation of SHM.
The initial position of the radius OP at t = 0 makes angle φ with the positive x-axis; after time t, OP makes angle (ωt + φ), so the x-projection gives the displacement at that instant.
The velocity of P′ equals the x-component of the velocity of P: v(t) = −ωA sin(ωt + φ); the acceleration of P′ equals the x-component of P's centripetal acceleration: a(t) = −ω²A cos(ωt + φ).
When projection is taken on the y-axis, the displacement is y(t) = A sin(ωt + φ), also SHM with the same amplitude A and angular frequency ω.
The y-projection differs from the x-projection by a phase of π/2; both have the same amplitude and period, but their phases are shifted by a quarter cycle.
Despite this geometric link, the force on a particle in linear SHM differs fundamentally from the centripetal force needed to maintain circular motion.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 5

Long Form

Analyse the force law for SHM using Newton's second law. Distinguish between a linear harmonic oscillator and a non-linear oscillator, and define the restoring force.

Applying Newton's second law to a particle of mass m executing SHM: F = ma = m(−ω²x) = −mω²x; defining k = mω², the force law becomes F(t) = −kx(t).
This force is always directed towards the mean position (opposite to displacement), hence called the restoring force; it vanishes at the mean position and is maximum at the extremes.
The angular frequency is related to k and m by ω = √(k/m), and the time period by T = 2π√(m/k); both depend only on system properties, not on amplitude.
A linear harmonic oscillator is one where the restoring force is strictly linearly proportional to displacement: F = −kx; examples include a mass-spring system obeying Hooke's law.
A non-linear oscillator has additional terms proportional to x², x³, etc., in the force; such systems do not execute perfect SHM and may display more complex periodic behaviour.
SHM can be defined in two equivalent ways: via the displacement equation x = A cos(ωt + φ) or via the force law F = −kx; one can be derived from the other by differentiating or integrating twice.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 6

Long Form

Compare the displacement, velocity, and acceleration plots for SHM as functions of time. Explain the phase relationships and identify their respective maxima and minima.

For φ = 0, the displacement is x(t) = A cos ωt, varying between −A and +A; it is maximum at t = 0 and again at t = T, and zero at t = T/4 and t = 3T/4.
Velocity v(t) = −ωA sin ωt varies between −ωA and +ωA; it is zero when x is maximum (t = 0, T) and maximum in magnitude when x = 0 (t = T/4, 3T/4) — so velocity leads displacement by π/2.
Acceleration a(t) = −ω²A cos ωt varies between −ω²A and +ω²A; it is always opposite to displacement (phase difference π) and maximum when displacement is maximum.
All three quantities are sinusoidal with the same period T, but differ in amplitude (A, ωA, ω²A respectively) and phase.
At the mean position (x = 0): speed is maximum (ωA) and acceleration is zero. At extreme positions (x = ±A): speed is zero and acceleration magnitude is maximum (ω²A).
The energy interpretation is consistent: maximum KE at x = 0 (maximum v) and maximum PE at x = ±A (maximum |x|), confirming conservation of total energy E = ½kA².

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 7

Long Form

Analyse why the total energy of a simple harmonic oscillator is proportional to the square of the amplitude, and discuss what happens to energy distribution at the mean position and extreme positions.

Total mechanical energy E = ½kA² is derived from E = K + U = ½kA²(sin² + cos²) = ½kA²; hence E is proportional to A² (quadratic, not linear).
Doubling the amplitude quadruples the total energy; this has practical significance in systems like musical instruments where loudness depends on amplitude of vibration.
At the mean position (x = 0): K = ½kA² (maximum), U = 0; all energy is kinetic — the particle is at its fastest and there is no elastic deformation.
At the extreme positions (x = ±A): K = 0, U = ½kA² (maximum); all energy is potential — the spring is maximally compressed or extended and the particle is momentarily at rest.
At an intermediate position x: K = ½k(A² − x²) and U = ½kx²; they add to give ½kA² irrespective of x, confirming conservation of total energy under conservative forces.
Both K and U peak twice per period T (period of each is T/2), confirming that energy exchange is twice as fast as the oscillation itself.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 8

Long Form

Explain periodic motion with the concept of period and frequency. Discuss the significance of Fourier's theorem in the context of periodic oscillations.

Periodic motion is one that repeats itself at regular intervals of time; the smallest such interval is the period T (in seconds). Frequency ν = 1/T is the number of repetitions per unit time, measured in hertz (Hz = s⁻¹).
The period of vibrations of a quartz crystal is expressed in microseconds (10⁻⁶ s), while the orbital period of Mercury is 88 earth days — illustrating the vast range of periodic phenomena.
Simple harmonic motion uses the simplest periodic functions: f(t) = A cos ωt or A sin ωt, with period T = 2π/ω; linear combinations such as A sin ωt + B cos ωt are also periodic with the same period.
Fourier's theorem, proved by Jean Baptiste Joseph Fourier (1768–1830), states that any periodic function can be expressed as a superposition of sine and cosine functions of different time periods with suitable coefficients.
This is profoundly important: complex oscillations in music, seismic waves, and electrical signals can all be decomposed into simple harmonic components using Fourier analysis.
Functions like e^(−ωt) and log(ωt) are non-periodic because they change monotonically and never repeat their values, confirming that not all functions of time represent physical oscillations.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 9

Long Form

With reference to a block-spring system, derive the time period of oscillation and explain the effect of attaching two identical springs on either side of the block.

For a block of mass m attached to a spring of constant k on a frictionless surface: when displaced by x, the spring exerts restoring force F = −kx; by Newton's second law, ma = −kx, giving a = −(k/m)x.
This matches the SHM equation a = −ω²x, so ω = √(k/m) and the time period is T = 2π√(m/k).
When two identical springs (each of constant k) are attached on both sides of the block: displacing the block by x elongates one spring by x and compresses the other by x, so each exerts force kx toward the mean position.
Net restoring force F = −kx − kx = −2kx; comparing with F = −k_eff x, the effective spring constant is k_eff = 2k.
The time period with two springs becomes T = 2π√(m/2k) = (1/√2) × 2π√(m/k), which is smaller (faster oscillation) than with a single spring.
This result demonstrates that the period depends on the effective restoring force constant, and that increasing stiffness (higher k_eff) reduces the period.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 10

Long Form

Discuss the amplitude, angular frequency, and phase constant as the three parameters characterising SHM. Explain how two SHMs can differ in each of these independently.

Amplitude A is the magnitude of maximum displacement; it is always positive and determines the extent of oscillation. Two SHMs with same ω and φ but different amplitudes A and B have identical shapes of motion but different extents.
Angular frequency ω = 2π/T = 2πν determines how fast the oscillation occurs; ω is the same for two SHMs that have the same period T. A SHM with larger ω oscillates faster (shorter T, higher ν).
Phase constant φ determines the initial state of the oscillation at t = 0; two SHMs with same A and ω but different φ are displaced in phase — the second effectively starts ahead or behind the first in the cycle.
Total energy E = ½kA² depends only on amplitude; changing φ or shifting the start time does not change the energy of the oscillation.
The period T = 2π/ω is independent of amplitude A, energy E, and phase constant φ — a fundamental property of SHM that distinguishes it from, say, planetary orbits (Kepler's third law).
To completely determine a linear SHM with known ω, two initial conditions are needed and sufficient — e.g., (i) initial position and velocity, (ii) amplitude and phase, or (iii) energy and phase.

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Chapter 13: Oscillations | Long Questions

Derive the expressions for velocity and acceleration of a particle in SHM using both the reference circle method and the differentiation method. Compare the results.

Derive an expression for the total mechanical energy of a particle executing SHM. Show that it is constant and analyse how energy is distributed between kinetic and potential forms.

Prove that the motion of a simple pendulum is simple harmonic for small displacements and derive the formula for its time period.

Explain how SHM can be viewed as the projection of uniform circular motion on a diameter. Describe what changes when the projection is taken on the y-axis instead of the x-axis.

Analyse the force law for SHM using Newton's second law. Distinguish between a linear harmonic oscillator and a non-linear oscillator, and define the restoring force.

Compare the displacement, velocity, and acceleration plots for SHM as functions of time. Explain the phase relationships and identify their respective maxima and minima.

Analyse why the total energy of a simple harmonic oscillator is proportional to the square of the amplitude, and discuss what happens to energy distribution at the mean position and extreme positions.

Explain periodic motion with the concept of period and frequency. Discuss the significance of Fourier's theorem in the context of periodic oscillations.

With reference to a block-spring system, derive the time period of oscillation and explain the effect of attaching two identical springs on either side of the block.

Discuss the amplitude, angular frequency, and phase constant as the three parameters characterising SHM. Explain how two SHMs can differ in each of these independently.

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