Simple Pendulum and Angular SHM

Simple Pendulum and Angular SHM

A pendulum oscillates with \( \theta_{\text{max}} = 0.2 \, \text{rad}, L = 2 \, \text{m}, g = 10 \, \text{m/s}^2 \). Wha

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. ω = √((g/L)) = √((10/2)) = √(5) ≈ 2.24 rad/s . Arc length amplitude: A = L θₘₐₓ = 2 × 0.2 = 0.4 m . vₘₐₓ = ω A = 2.24 × 0.4 ≈ 0.896 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x

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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.

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 1.69 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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 √((1.69/9.8)) ≈ 2 × 3.14 √(0.1724) ≈ 2.61 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.61 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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Simple Pendulum and Angular SHM

A simple pendulum has a period of \( 3 \, \text{s} \) on Earth. What will be its period on a planet where \( g = 2.45 \,

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. Period: T = 2π √((L/g)) . T ∝ (1/√(g)) . (Tₚlₐₙₑt/TEₐrth) = √((gEₐrth/gₚlₐₙₑt)) = √((9.8/2.45)) = √(4) = 2 . Tₚlₐₙₑt = 2 × 3 = 6 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 6 s

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Simple Pendulum and Angular SHM

What is a key reason the simple pendulum deviates from SHM at large angular displacements?

**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. At large angles, the sinusoidal restoring torque ( tau = -mgL sin θ ) introduces non-linear terms, disrupting the linear proportionality required for SHM. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring torque becomes non-linear follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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Simple Pendulum and Angular SHM

For a simple pendulum, why does the approximation of SHM break down at large amplitudes?

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. At large amplitudes, sin θ neq θ , and higher-order terms in the expansion ( sin θ = θ - (θ³/6) + ldots ) become significant, making the restoring force non-linear. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result

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Simple Pendulum and Angular SHM

The period of a simple pendulum is \( 2 \, \text{s} \) when \( g = 9.8 \, \text{m/s}^2 \). What should be the length of

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. T = 2π √((L/g)) . 2 = 2π √((L/9.8)) ⇒ 1 = π √((L/9.8)) . √((L/9.8)) = (1/π) ⇒ (L/9.8) = (1/π²) ⇒ L = (9.8/π²) ≈ 1 m (using π² ≈ 9.87 ). Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²

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Simple Pendulum and Angular SHM

Which of the following best explains why the period of a simple pendulum is independent of its mass?

**Pendulum motion** exhibits isochronism for small amplitudes, period depends only on L and g. Using g = 9.8 m/s², T calculation requires √(L/g), frequency reciprocal of period. Angular frequency directly √(g/L), e.g., L = 0.25 m gives T ≈ 1.0 s. The period T = 2π √((L/g)) depends on length and gravity. Mass cancels out in the equation of motion ( a = -(g/L) θ ), as both force and inertia scale with mass. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring force and inertia both depend on

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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

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Simple Pendulum and Angular SHM

Why does the period of a simple pendulum remain constant regardless of the bob’s material?

**Pendulum motion** exhibits isochronism for small amplitudes, period depends only on L and g. Using g = 9.8 m/s², T calculation requires √(L/g), frequency reciprocal of period. Angular frequency directly √(g/L), e.g., L = 0.25 m gives T ≈ 1.0 s. The period T = 2π √((L/g)) depends on length and gravity, not mass or material, as mass cancels out in the dynamics (force and inertia scale equally). Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The mass cancels out in the equation follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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Simple Pendulum and Angular SHM

A simple pendulum has a period of \( 1.4 \, \text{s} \) on Earth (\( g = 9.8 \, \text{m/s}^2 \)). What is its length?

**Simple pendulum** for small angles approximates SHM with period T = 2π√(L/g), frequency f = (1/2π)√(g/L), angular frequency ω = √(g/L) (rad/s), independent of mass. L length from pivot to centre of mass (m), g = 9.8 m/s² acceleration due to gravity, approximation sinθ ≈ θ (rad) for θ < 10°. T = 2π √((L/g)) . 1.4 = 2π √((L/9.8)) ⇒ √((L/9.8)) = (1.4/2π) ≈ 0.223 . (L/9.8) = (0.223)² ⇒ L ≈ 9.8 × 0.0497 ≈ 0.487 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.487 m

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 1.44 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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 √((1.44/9.8)) ≈ 2 × 3.14 √(0.1469) ≈ 2.406 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.406 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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