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#maximum speed

9 public questions tagged with this topic.

Forced Oscillations and Resonance

A particle’s x-projection from circular motion is \( x = 8 \cos (\pi t) \) (in m). What is its maximum speed?

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. Maximum speed: vₘₐₓ = ω A . A = 8 m, ω = π s⁻¹ . vₘₐₓ = π × 8 ≈ 3.14 × 8 ≈ 25.12 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 25.12 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

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Damped Oscillations

A particle’s x-projection from circular motion is \( x = 5 \cos (4t) \) (in m). What is its maximum speed?

**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. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 4 s⁻¹ . vₘₐₓ = 4 × 5 = 20 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 m/s 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 spring system has \( k = 800 \, \text{N/m}, m = 2 \, \text{kg} \). What is the maximum speed if amplitude is \( 5 \, \

**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)) = √((800/2)) = 20 rad/s . vₘₐₓ = ω A = 20 × 0.05 = 1.0 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.0 m/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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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

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

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Energy in SHM - Kinetic, Potential and Total

A particle’s x-projection from circular motion is \( x = 9 \cos (\pi t) \) (in m). What is its maximum speed?

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Maximum speed: vₘₐₓ = ω A . A = 9 m, ω = π s⁻¹ . vₘₐₓ = π × 9 ≈ 28.26 m/s . Applying x = A cos(ωt + φ), v =

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Simple Pendulum and Angular SHM

A pendulum bob oscillates with \( \theta = 0.1 \, \text{rad} \). If \( L = 1 \, \text{m}, g = 10 \, \text{m/s}^2 \), wha

**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. Maximum angular speed: ω = √((g/L)) = √((10/1)) = √(10) rad/s . Maximum speed: vₘₐₓ = ω A = ω (L θ) = √(10) × 1 × 0.1 ≈ 0.316 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.316 m/s follows, reflecting SHM

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

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Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 5 \cos (2t) \) (in m). What is its maximum speed?

**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. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 2 s⁻¹ . vₘₐₓ = 2 × 5 = 10 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 10 m/s

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

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Displacement, Velocity and Acceleration in SHM

A particle in circular motion has its x-projection as \( x = 7 \cos (\pi t) \) (in m). What is its maximum speed?

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Maximum speed: vₘₐₓ = ω A . A = 7 m, ω = π s⁻¹ . vₘₐₓ = π × 7 ≈ 3.14 × 7 ≈ 21.98 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20.0 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Straight Line Motion - Practice Set

A 1000kg car on a banked road (θ\=15∘, μs\=0.3) turns at radius 40m. What is the maximum speed without slipping? (Take g

Max speed vmax=rgμs+tan⁡θ1−μstan⁡θ. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 14 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0

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