Equations of SHM, Phase and Angular Frequency

Equations of SHM, Phase and Angular Frequency

A spring-mass system has \( m = 1.25 \, \text{kg}, k = 500 \, \text{N/m} \). What is its angular frequency?

**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. ω = √((k/m)) = √((500/1.25)) = √(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.

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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.

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Equations of SHM, Phase and Angular Frequency

A pendulum oscillates with a period of \( 1.5 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**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. T = 2π √((L/g)) . 1.5 = 2π √((L/9.8)) ⇒ (1.5/2π) = √((L/9.8)) . ((1.5/2 × 3.14))² = (L/9.8) ⇒ L = 9.8 × (0.238)² ≈ 0.56 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.56

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Equations of SHM, Phase and Angular Frequency

What happens to the frequency of a spring-mass system if the spring constant is quadrupled and the mass is halved?

**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. Frequency v = (1/2π) √((k/m)) . If k' = 4k and m' = (m/2) , then v' = (1/2π) √((4k/m/2)) = (1/2π) √((8k/m)) = √(8) · v = 2√(2) · v , doubling the original frequency by √(2) . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

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Equations of SHM, Phase and Angular Frequency

A simple pendulum has a frequency of \( 0.25 \, \text{Hz} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its lengt

**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°. Period: T = (1/v) = (1/0.25) = 4 s . T = 2π √((L/g)) ⇒ 4 = 2π √((L/9.8)) . √((L/9.8)) = (4/2π) ≈ 0.637 ⇒ (L/9.8) = (0.637)² ⇒ L ≈ 3.98 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.98 m follows, reflecting

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Equations of SHM, Phase and Angular Frequency

Which condition must be satisfied for the potential energy in an SHM system to be expressible as a quadratic function of

**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. Potential energy in SHM ( U = (1/2) k x² ) is quadratic when the force is conservative and linear ( F = -kx ), distinguishing SHM from non-linear systems. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

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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.

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Equations of SHM, Phase and Angular Frequency

Which of the following is a non-periodic motion? (\( \omega \) is a positive constant)

**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°. (a) sin ω t + cos ω t : Periodic, period (2π/ω) . (b) sin³ ω t : Periodic, period (2π/3ω) . (c) e⁻ω t : Non-periodic (decays to zero). (d) cos 2ω t : Periodic, period (π/ω) . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

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Equations of SHM, Phase and Angular Frequency

A spring system has \( m = 0.3 \, \text{kg}, k = 120 \, \text{N/m}, A = 8 \, \text{cm} \). What is the potential energy

**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. Potential energy: U = (1/2) k x² . k = 120 N/m, x = 0.04 m . U = 0.5 × 120 × (0.04)² = 0.5 × 120 × 0.0016 = 0.096 J . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =

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Equations of SHM, Phase and Angular Frequency

Two identical springs (\( k = 80 \, \text{N/m} \)) are attached to a \( 2 \, \text{kg} \) mass as in Fig. 13.14. What is

**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 × 80 = 160 N/m . T = 2π √((m/kₑff)) = 2π √((2/160)) = 2π √(0.0125) ≈ 0.702 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.702 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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Equations of SHM, Phase and Angular Frequency

A spring-mass system has \( m = 1.6 \, \text{kg}, k = 640 \, \text{N/m} \). What is its angular frequency?

**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°. ω = √((k/m)) = √((640/1.6)) = √(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 > Equations of SHM, Phase and Angular Frequency

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Equations of SHM, Phase and Angular Frequency

In SHM, what condition results in the potential energy being zero?

**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. Potential energy U = (1/2) k x² is zero when displacement x = 0 , which occurs at the mean position, where all energy is kinetic. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result Zero displacement follows, reflecting

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