Chapter 13: Oscillations Application Questions | physics Class 11 CBSE

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A child swinging on a playground swing notices that the time for one complete back-and-forth motion remains the same whether she swings a little or a lot. With reference to the simple pendulum model, explain this observation.

The swing behaves as a simple pendulum; its time period T = 2π√(L/g) depends only on the effective length L of the swing and local g, not on the amplitude of swinging.
For small angular displacements, sin θ ≈ θ, and the restoring torque is linearly proportional to displacement, making the motion SHM with a period independent of amplitude.
When the child swings more (larger amplitude), the restoring torque is stronger and the bob moves faster, but the period remains the same — speed and amplitude change together in a way that leaves T unchanged.
This amplitude-independence of period is a fundamental property of SHM; it fails for very large amplitudes where sin θ can no longer be approximated by θ and the motion is no longer strictly SHM.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

A piston in a steam engine moves back and forth executing SHM. If its stroke (twice the amplitude) is 1.0 m and angular frequency is 200 rad/min, calculate its maximum speed.

The stroke is twice the amplitude, so amplitude A = 1.0/2 = 0.5 m.
Maximum speed in SHM occurs at the mean position and equals v_max = ωA.
v_max = 200 rad/min × 0.5 m = 100 m/min = 100/60 m/s ≈ 1.67 m/s.
This maximum speed is achieved twice per cycle (once in each direction) as the piston passes through the centre of its stroke, where all energy is kinetic.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 3

Applied Concept

A geologist studying seismic waves notes that Earth can be modelled as a system of coupled oscillators. Explain, with reference to the chapter, how oscillatory motion of particles within Earth leads to wave propagation.

Any material medium, including the Earth, can be modelled as a collection of a large number of coupled oscillators — atoms or groups of atoms connected through interatomic forces.
When one oscillator is disturbed (as in an earthquake), it disturbs neighbouring oscillators through coupling; the disturbance propagates outward as a wave — in this case, seismic waves.
The collective oscillations of the medium's constituents manifest as waves; examples given in the chapter include water waves, seismic waves, and electromagnetic waves — all arising from coupled oscillatory motion.
The energy of the disturbance travels outward at a speed determined by the elastic properties of the medium, without the medium itself translating; each particle oscillates locally while the wave pattern propagates.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 4

Applied Concept

A body attached to a spring (k = 50 N/m, m = 1 kg) is pulled to x = 10 cm and released from rest. Calculate the kinetic, potential, and total energy when the body is 5 cm from the mean position.

Angular frequency ω = √(k/m) = √(50/1) = 7.07 rad/s; at x = 5 cm = 0.05 m, cos(ωt) = 0.5, so sin(ωt) = √3/2 = 0.866.
Velocity at x = 0.05 m: v = ωA × sin(ωt) = 7.07 × 0.1 × 0.866 = 0.612 m/s; KE = ½ × 1 × (0.612)² ≈ 0.19 J.
PE at x = 0.05 m: U = ½kx² = ½ × 50 × (0.05)² = 0.0625 J.
Total energy = KE + PE = 0.19 + 0.0625 = 0.25 J; this equals ½kA² = ½ × 50 × (0.1)² = 0.25 J, confirming energy conservation.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A musician tunes her guitar by tightening a string. With reference to the force law for SHM and the concept of frequency, explain why tightening increases the pitch (frequency) of the sound produced.

The vibrating string in a guitar acts as an oscillator; tightening the string increases the tension, which acts like increasing the restoring force constant k in the analogy F = −kx.
From the force law: ω = √(k/m); as the effective k increases with tighter tension, ω increases and hence frequency ν = ω/2π increases.
Higher frequency means more oscillations per second, which the ear perceives as a higher pitch; doubling k increases frequency by a factor of √2.
This is directly analogous to the spring-mass system where ω = √(k/m): a stiffer spring (larger k) oscillates faster, just as a tighter string vibrates faster.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 6

Applied Concept

A spring balance reads up to 50 kg over a scale length of 20 cm. A body attached to it oscillates with period 0.6 s. Find the weight of the body.

Spring constant k: maximum force = 50 × 9.8 = 490 N; maximum extension = 20 cm = 0.2 m; so k = 490/0.2 = 2450 N/m.
Time period T = 2π√(m/k); squaring and rearranging: m = k × T²/(4π²) = 2450 × (0.6)²/(4π²).
m = 2450 × 0.36/39.48 = 882/39.48 ≈ 22.3 kg.
Weight of the body = mg = 22.3 × 9.8 ≈ 219 N.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

An astronaut on the Moon uses a simple pendulum to measure local gravity. If the pendulum of length 1 m has a period of 4.8 s on the Moon, calculate the acceleration due to gravity on the Moon's surface.

From the pendulum formula T = 2π√(L/g), solving for g: g = 4π²L/T².
Substituting L = 1 m and T = 4.8 s: g = 4π² × 1/(4.8)² = 39.48/23.04 ≈ 1.71 m/s².
This matches the Moon's surface gravity of 1.7 m/s² given in the chapter, confirming the validity of the pendulum formula.
This method — measuring the period of a pendulum of known length — is a classic technique for determining g at any location, illustrating the practical application of SHM.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

A body executes SHM described by x = 5 cos[2πt + π/4] (SI units). At t = 1.5 s, find its displacement, speed, and acceleration.

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 × 0.707 = −3.535 m.
Speed: v = −5 × 2π × sin(3π + π/4) = −10π × sin(3π + π/4) = 10π × sin(π/4) = 10π × 0.707 ≈ 22.2 m/s (magnitude).
Acceleration: a = −ω²x = −(2π)² × (−3.535) = 4π² × 3.535 ≈ 139.6 m/s² ≈ 140 m/s² (directed toward mean position).
These results confirm the relationships v = −ωA sin(ωt + φ) and a = −ω²x; the negative displacement at this instant means acceleration is positive (directed toward x = 0).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 9

Applied Concept

A cork floating in a liquid is pushed down and released; it bobs up and down. With reference to the chapter, explain the physics of this motion and state why it is SHM.

When the cork is pushed down by a small distance x from its equilibrium (floating) position, the buoyant force increases while the weight stays constant, producing a net upward restoring force.
The restoring force is proportional to x: F = −ρ_l g A x (where A is cross-sectional area and ρ_l is liquid density), which has the form F = −kx, satisfying the SHM force law.
Comparing with F = −mω²x: the angular frequency is ω = √(ρ_l g A / m) and the time period is T = 2π√(hρ/ρ_l g), where h is the height of the cork and ρ is its density.
This is the example given in Exercise 13.17 of the chapter; the cork executes SHM as long as damping due to viscosity of the liquid is ignored.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

In designing a clock using a simple pendulum, an engineer wants the pendulum to tick every second (T = 2 s). What length should the pendulum have? (g = 9.8 m/s²)

From T = 2π√(L/g), solving for L: L = gT²/4π².
Substituting T = 2 s and g = 9.8 m/s²: L = 9.8 × 4/(4π²) = 39.2/39.48 ≈ 1 m.
The pendulum must be approximately 1 m long to tick every second; this is called a seconds pendulum.
If g is different (e.g., at a different altitude or on a different planet), the required length changes proportionally to g; this is why pendulum clocks must be calibrated for their location.

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Chapter 13: Oscillations | Application Questions

A child swinging on a playground swing notices that the time for one complete back-and-forth motion remains the same whether she swings a little or a lot. With reference to the simple pendulum model, explain this observation.

A piston in a steam engine moves back and forth executing SHM. If its stroke (twice the amplitude) is 1.0 m and angular frequency is 200 rad/min, calculate its maximum speed.

A geologist studying seismic waves notes that Earth can be modelled as a system of coupled oscillators. Explain, with reference to the chapter, how oscillatory motion of particles within Earth leads to wave propagation.

A body attached to a spring (k = 50 N/m, m = 1 kg) is pulled to x = 10 cm and released from rest. Calculate the kinetic, potential, and total energy when the body is 5 cm from the mean position.

A musician tunes her guitar by tightening a string. With reference to the force law for SHM and the concept of frequency, explain why tightening increases the pitch (frequency) of the sound produced.

A spring balance reads up to 50 kg over a scale length of 20 cm. A body attached to it oscillates with period 0.6 s. Find the weight of the body.

An astronaut on the Moon uses a simple pendulum to measure local gravity. If the pendulum of length 1 m has a period of 4.8 s on the Moon, calculate the acceleration due to gravity on the Moon's surface.

A body executes SHM described by x = 5 cos[2πt + π/4] (SI units). At t = 1.5 s, find its displacement, speed, and acceleration.

A cork floating in a liquid is pushed down and released; it bobs up and down. With reference to the chapter, explain the physics of this motion and state why it is SHM.

In designing a clock using a simple pendulum, an engineer wants the pendulum to tick every second (T = 2 s). What length should the pendulum have? (g = 9.8 m/s²)

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