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#frequency

89 public questions tagged with this topic.

Forced Oscillations and Resonance

A particle in SHM has \( a = -36 x \) (in SI units). What is its frequency?

**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. For SHM, a = -ω² x . Given a = -36 x , ω² = 36 ⇒ ω = 6 rad/s . Frequency: v = (ω/2π) = (6/2 × 3.14) ≈ 0.955 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.955 Hz follows, reflecting SHM dependence

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Forced Oscillations and Resonance

A pendulum of length \( 0.36 \, \text{m} \) oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its frequency?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. Period: T = 2π √((L/g)) = 2π √((0.36/9.8)) ≈ 2 × 3.14 √(0.0367) ≈ 1.2 s . Frequency: v = (1/T) = (1/1.2) ≈ 0.833 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.833 Hz follows, reflecting SHM dependence on

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Damped Oscillations

Which feature of SHM explains why two particles with identical amplitude and frequency may not reach their extreme posit

**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. Different phase constants ( Φ ) shift the oscillation cycles, causing particles to reach extremes at different times despite equal amplitude and frequency. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Difference in phase constants 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 particle in SHM has \( a = -16 x \) (in SI units). What is its frequency?

**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. For SHM, a = -ω² x . Given a = -16 x , ω² = 16 ⇒ ω = 4 rad/s . Frequency: v = (ω/2π) = (4/2 × 3.14) ≈ 0.637 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.637 Hz

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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Equations of SHM, Phase and Angular Frequency

A particle in SHM has an amplitude of \( 8 \, \text{cm} \) and a frequency of \( 2 \, \text{Hz} \). What is its maximum

**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. Maximum acceleration: aₘₐₓ = ω² A . ω = 2π v = 2 × 3.14 × 2 = 12.56 rad/s . A = 8 cm = 0.08 m . aₘₐₓ = (12.56)² × 0.08 ≈ 157.75 × 0.08 ≈ 12.62 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²

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

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Simple Pendulum and Angular SHM

A pendulum of length \( 2 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its frequency?

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = 2π √((L/g)) = 2π √((2/10)) = 2π √(0.2) ≈ 2.8 s . Frequency: v = (1/T) = (1/2.8) ≈ 0.357 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.357 Hz follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM

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Spring-Mass System and Combination of Springs

Two identical springs (\( k = 75 \, \text{N/m} \)) are attached to a \( 1.5 \, \text{kg} \) mass as in Fig. 13.14. What

**Oscillation of mass attached to spring** follows Hooke's law F = -k x, SHM with ω = √(k/m). Effective stiffness for two identical springs in parallel doubles, k_eff = 2k, increasing frequency by √2, while series halves stiffness to k/2, lowering frequency. Energy E = ½ k_eff A². Effective kₑff = 2k = 2 × 75 = 150 N/m . ω = √((kₑff/m)) = √((150/1.5)) = √(100) = 10 rad/s . v = (ω/2π) = (10/2 × 3.14) ≈ 1.59 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Simple Pendulum and Angular SHM

A simple pendulum has a frequency of \( 0.4 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = (1/v) = (1/0.4) = 2.5 s . T = 2π √((L/g)) ⇒ 2.5 = 2π √((L/9.8)) . √((L/9.8)) = (2.5/2π) ≈ 0.398 ⇒ (L/9.8) = (0.398)² ⇒ L ≈ 1.55 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.55 m

Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( a = -81 x \) (in SI units). What is its frequency?

**Distinction between periodic and oscillatory** clarifies all SHM is periodic but not all periodic is SHM. SHM requires linear restoring force and inertia, a ∝ -x, with ω = √(k/m). Functions like sin²ωt have period π/ω but lack a = -ω² x, thus periodic not SHM, while uniform circular motion is periodic without linear oscillation. For SHM, a = -ω² x . Given a = -81 x , ω² = 81 ⇒ ω = 9 rad/s . Frequency: v = (ω/2π) = (9/2 × 3.14) ≈ 1.43 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Spring-Mass System and Combination of Springs

A spring-mass system has \( m = 1.8 \, \text{kg}, k = 720 \, \text{N/m} \). What is its frequency?

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Angular frequency: ω = √((k/m)) = √((720/1.8)) = √(400) = 20 rad/s . Frequency: v = (ω/2π) = (20/2 × 3.14) ≈ 3.18 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.18 Hz follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Simple Pendulum and Angular SHM

A pendulum of length \( 1.0 \, \text{m} \) oscillates with \( g = 10 \, \text{m/s}^2 \). What is its frequency?

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = 2π √((L/g)) = 2π √((1/10)) ≈ 2 × 3.14 √(0.1) ≈ 1.986 s . Frequency: v = (1/T) = (1/1.986) ≈ 0.503 Hz . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.503 Hz follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM

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