Chapter 13: Oscillations Case Study Questions | physics Class 11 CBSE

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

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Riya is conducting a laboratory experiment on a spring-mass system. She fixes one end of a spring (k = 200 N/m) to a wall and attaches a block of mass 2 kg to the free end on a frictionless surface. She pulls the block 5 cm from its equilibrium position and releases it from rest. She notices that the block oscillates back and forth smoothly, always returning to the same extreme positions. She records the time for 20 complete oscillations as 12.57 seconds. Her teacher asks her to analyse the displacement, velocity, and acceleration of the block at various positions during its oscillation, and to verify the energy conservation principle by computing the total mechanical energy at different points in the motion.

Question 1: What is the time period and angular frequency of the block's oscillation in Riya's experiment?

Time period T = total time / number of oscillations = 12.57 / 20 = 0.628 s, which matches T = 2π√(m/k) = 2π√(2/200) = 2π × 0.1 = 0.628 s.
Angular frequency ω = 2π/T = 2π/0.628 = 10 rad/s, or equivalently ω = √(k/m) = √(200/2) = √100 = 10 rad/s.
Both approaches give the same result, confirming the formula ω = √(k/m) for a spring-mass system executing SHM.

Question 2: Calculate the maximum speed and maximum acceleration of the block in Riya's experiment.

Maximum speed occurs at the mean position (x = 0): v_max = ωA = 10 × 0.05 = 0.5 m/s.
Maximum acceleration occurs at the extreme positions (x = ±A): a_max = ω²A = 100 × 0.05 = 5 m/s².
At the mean position acceleration is zero, and at the extreme positions speed is zero — confirming the phase difference of π/2 between velocity and displacement.

Question 3: Verify energy conservation in Riya's experiment by computing total energy and comparing kinetic and potential energies when the block is 3 cm from the mean position.

Total energy E = ½kA² = ½ × 200 × (0.05)² = ½ × 200 × 0.0025 = 0.25 J.
At x = 0.03 m: PE = ½kx² = ½ × 200 × (0.03)² = 0.09 J; KE = E − PE = 0.25 − 0.09 = 0.16 J.
Speed at x = 0.03 m: v = ω√(A² − x²) = 10 × √(0.0025 − 0.0009) = 10 × √0.0016 = 10 × 0.04 = 0.4 m/s; KE = ½mv² = ½ × 2 × 0.16 = 0.16 J, consistent with above.

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Case Set 2

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A physics teacher sets up a demonstration using a simple pendulum of length 1 m in the classroom (g = 9.8 m/s²). She displaces the bob by a small angle of 5° and releases it. Students observe the smooth to-and-fro motion and note that the period remains constant regardless of whether the bob is displaced 3° or 8°. The teacher then challenges students to predict what would happen to the period if the same pendulum were taken to the surface of the Moon, where the acceleration due to gravity is 1.7 m/s². She also asks them to calculate the length of a pendulum on the Moon that would have the same period as the Earth pendulum.

Question 1: Calculate the time period of the pendulum in the classroom and state why the period does not change when the displacement angle is changed from 3° to 8°.

Time period T = 2π√(L/g) = 2π√(1/9.8) = 2π × 0.319 = 2.007 ≈ 2 s.
The period is independent of amplitude (displacement angle) for small angles because the approximation sin θ ≈ θ holds, making the restoring torque linearly proportional to θ — a condition for SHM.
Since T = 2π√(L/g) contains neither the amplitude nor the mass of the bob, changing the displacement from 3° to 8° does not affect the period.

Question 2: What would be the time period of the same pendulum (L = 1 m) on the surface of the Moon?

On the Moon: T_M = 2π√(L/g_M) = 2π√(1/1.7) = 2π × 0.767 = 4.82 s.
The period increases because Moon's gravity (1.7 m/s²) is weaker than Earth's (9.8 m/s²), making the restoring force smaller and thus slowing the oscillation.
The ratio T_M/T_E = √(g_E/g_M) = √(9.8/1.7) ≈ 2.4, so the Moon period is about 2.4 times the Earth period.

Question 3: Find the length of a simple pendulum on the Moon that would have the same time period as the 1 m pendulum on Earth (T = 2 s). What does this result reveal about the relationship between L and g for equal periods?

For the same period T = 2 s on Moon: T = 2π√(L_M/g_M), so L_M = g_M × T²/4π² = 1.7 × 4/(4π²) = 6.8/39.48 ≈ 0.172 m.
The Moon pendulum must be only 17.2 cm long to tick at the same rate as the 1 m Earth pendulum — much shorter, because weaker gravity requires shorter length to maintain the same restoring effect.
The result reveals that for equal periods, L/g = constant; on any planet, a seconds pendulum (T = 2 s) has length L = g/π², so L is directly proportional to g.

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Case Set 3

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During a physics practical, Ankit observes a ball tied to a string rotating uniformly in a horizontal circle of radius 0.3 m with an angular speed of 4 rad/s. His friend Priya, standing to the side, watches only the shadow of the ball on a vertical wall illuminated by a lamp directly above. Priya notices that the shadow moves back and forth along a straight line in a smooth repetitive manner. Their teacher explains that this is the fundamental connection between uniform circular motion and simple harmonic motion, and that the projection of circular motion on any diameter of the circle executes SHM. The teacher asks both students to derive the SHM parameters from the circular motion data.

Question 1: What is the amplitude, angular frequency, and time period of the SHM executed by the shadow on the wall?

The amplitude of the SHM equals the radius of the circular motion: A = 0.3 m.
The angular frequency of the SHM equals the angular speed of the circular motion: ω = 4 rad/s.
Time period T = 2π/ω = 2π/4 = π/2 ≈ 1.57 s.

Question 2: Write the displacement equation for the shadow's SHM if at t = 0 the ball is at the rightmost point of the circle (angle φ = 0 with x-axis).

At t = 0, the ball is at angle φ = 0 (rightmost point); the x-projection gives x(0) = A cos(0) = A = 0.3 m, which is the maximum positive displacement.
The displacement equation is x(t) = A cos(ωt + φ) = 0.3 cos(4t + 0) = 0.3 cos(4t) metres.
The velocity of the shadow is v(t) = −ωA sin(ωt) = −1.2 sin(4t) m/s, and acceleration is a(t) = −ω²x = −16 × 0.3 cos(4t) = −4.8 cos(4t) m/s².

Question 3: If the projection is now taken on the y-axis instead of the x-axis (with the same initial conditions), write the new displacement equation and explain the phase relationship between the two projections.

The y-projection of the same circular motion gives y(t) = A sin(ωt + φ) = 0.3 sin(4t) m, since the y-component of a position at angle ωt from the x-axis is A sin(ωt).
Comparing x(t) = 0.3 cos(4t) and y(t) = 0.3 sin(4t): both are SHMs with identical amplitude A = 0.3 m and angular frequency ω = 4 rad/s, but the y-projection leads the x-projection by a phase of π/2.
This π/2 phase difference arises because sin(ωt) = cos(ωt − π/2); the two projections on perpendicular diameters always differ by exactly π/2 regardless of amplitude or frequency.

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A body oscillates according to the equation x = 5 cos(2πt + π/4) metres, where t is in seconds. A student is asked to analyse this SHM completely. She identifies the amplitude as 5 m, angular frequency as 2π rad/s, and the phase constant as π/4. At t = 1.5 s, she is required to calculate the exact displacement, speed, and acceleration of the body. She also needs to determine the total energy and find the position at which kinetic energy equals potential energy, understanding that such calculations are typical in board examinations and require careful application of the SHM formulae for displacement, velocity, and acceleration.

Question 1: Identify the amplitude, angular frequency, time period, and phase constant of this SHM.

Amplitude A = 5 m (coefficient of the cosine function); angular frequency ω = 2π rad/s (coefficient of t inside the argument).
Time period T = 2π/ω = 2π/2π = 1 s; frequency ν = 1/T = 1 Hz.
Phase constant φ = π/4 rad, representing the initial phase at t = 0; initial displacement x(0) = 5 cos(π/4) = 5/√2 ≈ 3.54 m.

Question 2: Calculate the displacement and speed of the body at t = 1.5 s.

At t = 1.5 s: phase = 2π × 1.5 + π/4 = 3π + π/4; displacement x = 5 cos(3π + π/4) = −5 cos(π/4) = −5/√2 ≈ −3.54 m.
Speed: |v| = |−ωA sin(ωt + φ)| = 2π × 5 × |sin(3π + π/4)| = 10π × sin(π/4) = 10π/√2 ≈ 22.2 m/s.
The negative displacement indicates the body is on the negative side of the mean position at t = 1.5 s, while the high speed reflects it is near the mean position where KE is maximum.

Question 3: If the spring constant is k = 4π² N/m, find the total energy and determine at what displacement from the mean position KE equals PE.

Total energy E = ½kA² = ½ × 4π² × 25 = 50π² ≈ 493 J; alternatively E = ½mω²A² = ½ × 1 × (2π)² × 25 = 50π² J (mass m = k/ω² = 4π²/4π² = 1 kg).
For KE = PE: ½kA² = 2 × ½kx², so x² = A²/2, giving x = ±A/√2 = ±5/√2 ≈ ±3.54 m.
At x = ±3.54 m, KE = PE = ½ × total energy = 25π² ≈ 246.5 J; this position is not the midpoint (±2.5 m) — energy equality occurs at x = A/√2, not A/2.

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Case Set 5

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Meera is investigating the energy distribution in a spring-mass oscillator as part of her project. She uses a block of mass 1 kg attached to a spring of constant 50 N/m. The block is pulled to 10 cm from equilibrium and released from rest. She plots graphs of kinetic energy K, potential energy U, and total energy E as functions of both time and displacement. She notes that both K and U peak twice in each oscillation cycle, while the total energy remains a horizontal line. At a position 5 cm from the mean, she calculates all three energies and verifies the conservation principle, relating her findings to the work-energy theorem and conservative forces.

Question 1: Why do kinetic and potential energies each peak twice per oscillation period T, while displacement peaks only once per period?

Kinetic energy K = ½kA² sin²(ωt + φ) and potential energy U = ½kA² cos²(ωt + φ) both contain squared trigonometric functions, which have period π (not 2π).
Since sin²θ and cos²θ repeat every π radians, both K and U complete one full cycle in time T/2, peaking twice in each period T of the oscillation.
Displacement x = A cos(ωt + φ) completes one cycle in T; K depends on v² (always positive), so it is the same for +v and −v, causing two peaks per oscillation.

Question 2: Calculate the total energy and the kinetic and potential energies when the block is 5 cm from the mean position.

Total energy E = ½kA² = ½ × 50 × (0.10)² = 0.25 J; this is constant at all positions.
At x = 0.05 m: PE = ½kx² = ½ × 50 × (0.05)² = 0.0625 J; KE = E − PE = 0.25 − 0.0625 = 0.1875 J ≈ 0.19 J.
Speed at x = 0.05 m: v = ω√(A²−x²) = 7.07 × √(0.01−0.0025) = 7.07 × 0.0866 ≈ 0.612 m/s; KE = ½ × 1 × (0.612)² ≈ 0.187 J ✓.

Question 3: Explain why the spring force F = −kx is called a conservative force, and how this property is essential for the constancy of total mechanical energy in SHM.

A conservative force is one for which the work done depends only on the initial and final positions and not on the path taken; the spring force F = −kx satisfies this because it is derivable from a potential energy function U = ½kx².
Since no energy is lost to the path (unlike friction, which is non-conservative), the work done by the spring in a complete cycle is zero and mechanical energy is perfectly conserved.
The total energy E = K + U = ½kA² remains constant at all times and positions in ideal SHM; this constancy is only possible because the restoring spring force is conservative — any dissipative (non-conservative) force would reduce E over time.

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Chapter 13: Oscillations | Case Study Questions

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Question 1: What is the time period and angular frequency of the block's oscillation in Riya's experiment?

Question 2: Calculate the maximum speed and maximum acceleration of the block in Riya's experiment.

Question 3: Verify energy conservation in Riya's experiment by computing total energy and comparing kinetic and potential energies when the block is 3 cm from the mean position.

Passage

Question 1: Calculate the time period of the pendulum in the classroom and state why the period does not change when the displacement angle is changed from 3° to 8°.

Question 2: What would be the time period of the same pendulum (L = 1 m) on the surface of the Moon?

Question 3: Find the length of a simple pendulum on the Moon that would have the same time period as the 1 m pendulum on Earth (T = 2 s). What does this result reveal about the relationship between L and g for equal periods?

Passage

Question 1: What is the amplitude, angular frequency, and time period of the SHM executed by the shadow on the wall?

Question 2: Write the displacement equation for the shadow's SHM if at t = 0 the ball is at the rightmost point of the circle (angle φ = 0 with x-axis).

Question 3: If the projection is now taken on the y-axis instead of the x-axis (with the same initial conditions), write the new displacement equation and explain the phase relationship between the two projections.

Passage

Question 1: Identify the amplitude, angular frequency, time period, and phase constant of this SHM.

Question 2: Calculate the displacement and speed of the body at t = 1.5 s.

Question 3: If the spring constant is k = 4π² N/m, find the total energy and determine at what displacement from the mean position KE equals PE.

Passage

Question 1: Why do kinetic and potential energies each peak twice per oscillation period T, while displacement peaks only once per period?

Question 2: Calculate the total energy and the kinetic and potential energies when the block is 5 cm from the mean position.

Question 3: Explain why the spring force F = −kx is called a conservative force, and how this property is essential for the constancy of total mechanical energy in SHM.

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