#oscillation

Torque on Magnetic Dipole and Potential Energy

A magnetic dipole oscillates in a uniform field when displaced from its equilibrium position because:

**Potential energy of magnetic dipole** U = -m B cosθ explains stability. Given m = 0.9 A·m², B = 0.5 T, θ = 90°, sin90° = 1, τ = 0.45 N·m. For 60°, sin60° = √3/2 ≈0.866, reducing torque proportionally. The torque on a magnetic dipole ( tau = m B sinθ ) acts as a restoring force when displaced from its equilibrium position (aligned with the field). This torque causes oscillatory motion, similar to a pendulum, as it seeks to return to the stable alignment. Substituting values gives The torque acts as a restoring force, which matches expected magnitude for this magnetic configuration, confirming

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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Damped Oscillations

Two identical springs (\( k = 100 \, \text{N/m} \)) are attached to a \( 2.0 \, \text{kg} \) mass as in Fig. 13.14. What

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Effective kₑff = 2k = 2 × 100 = 200 N/m . T = 2π √((m/kₑff)) = 2π √((2/200)) = 2π √(0.01) = 2π × 0.1 ≈ 0.628 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.628 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A spring-mass system has \( m = 2 \, \text{kg}, k = 800 \, \text{N/m} \). What is its angular frequency?

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. ω = √((k/m)) = √((800/2)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 rad/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A spring-mass system oscillates with \( T = 0.4 \, \text{s} \) when \( m = 0.2 \, \text{kg} \). What is the spring const

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. T = 2π √((m/k)) . 0.4 = 2π √((0.2/k)) ⇒ (0.4/2π) = √((0.2/k)) . (0.0637)² = (0.2/k) ⇒ k = (0.2/0.00406) ≈ 49.26 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 49.26 N/m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A spring-mass system oscillates with \( T = 0.6 \, \text{s} \) when \( m = 0.9 \, \text{kg} \). What is the spring const

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. T = 2π √((m/k)) . 0.6 = 2π √((0.9/k)) ⇒ (0.6/2π) = √((0.9/k)) . (0.0955)² = (0.9/k) ⇒ k = (0.9/0.00912) ≈ 98.68 N/m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 98.68 N/m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

Which function represents SHM? (\( \omega \) is a positive constant)

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. SHM requires a = -ω² x : (a) 6 sin (2ω t - (π/4)) : a = -6 (2ω)² sin (2ω t - (π/4)) = -ω² x , SHM. (b) sin ω t + cos 3ω t : Not SHM (mixed frequencies). (c) eω t : Not periodic. (d) cos³ ω t : Periodic, not SHM. Applying x = A cos(ωt +

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

What is the effect on the frequency of a simple pendulum if it is taken to a planet where gravity is one-fourth that of

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. Frequency v = (1/2π) √((g/L)) . If g' = (g/4) , then v' = (1/2π) √((g/4/L)) = (1/2) v , halving the frequency. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It halves follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

Which function represents simple harmonic motion? (\( \omega \) is a positive constant)

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. SHM requires a = -ω² x : (a) cos ω t + sin 2ω t : Not SHM (mixed frequencies). (b) 4 sin (ω t + (π/6)) : a = -4ω² sin (ω t + (π/6)) = -ω² x , SHM. (c) eω t : Not periodic. (d) sin² ω t : Periodic, not SHM. Applying x = A cos(ωt + φ),

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

What distinguishes the oscillatory nature of a pendulum from the periodic motion of a planet in orbit?

**Damped oscillations** occur when resistive forces dissipate energy, amplitude decays exponentially as A(t) = A₀ e^(-b t/2m), b damping coefficient (kg/s), frequency slightly reduced ω' = √(ω₀² - (b/2m)²). Damping arises from friction or viscosity, energy loss per cycle proportional to velocity squared, motion eventually stops. A pendulum oscillates about a fixed point due to a linear restoring force (gravity component), while a planet’s orbit is periodic but governed by inverse-square gravitational force. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring force is linear follows, reflecting SHM dependence

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Equations of SHM, Phase and Angular Frequency

Which statement best explains why uniform circular motion is not considered oscillatory despite being periodic?

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Oscillatory motion requires to-and-fro movement about an equilibrium, while uniform circular motion involves continuous rotation without reversing direction. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It does not involve to-and-fro motion follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Equations of SHM, Phase and Angular Frequency

Two identical springs (\( k = 50 \, \text{N/m} \)) are attached to a \( 0.5 \, \text{kg} \) mass as in Fig. 13.14. What

**SHM representation** using sine or cosine equivalent with phase offset, ω relates to system parameters like mass and stiffness. Understanding ω and φ permits prediction of position at any time and comparison of two SHM via phase difference Δφ = φ₂ - φ₁. Effective kₑff = 2k = 2 × 50 = 100 N/m . T = 2π √((m/kₑff)) = 2π √((0.5/100)) = 2π √(0.005) ≈ 0.44 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.44 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Equations of SHM, Phase and Angular Frequency

What fundamental property of the restoring force distinguishes simple harmonic motion from other oscillatory motions?

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. In SHM, the restoring force must be directly proportional to displacement and opposite in direction ( F = -kx ), ensuring a linear relationship, unlike non-linear oscillatory systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Its linearity

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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