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#speed calculation

15 public questions tagged with this topic.

Forced Oscillations and Resonance

A particle in SHM has \( x = 4 \cos (2t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 2 s⁻¹, Φ = (π/3) . At t = 0.5 : 2 × 0.5 + (π/3) = 1 + (π/3) ≈ 2.047 rad ≈ 117° . v = -2 × 4 sin 117° ≈ -8 sin (180° - 63°) ≈ -8 × 0.838 ≈ -6.7 m/s

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

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Forced Oscillations and Resonance

A particle in SHM has \( x = 4 \sin (4t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)? (Ta

**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. Velocity: v = ω A cos (ω t + Φ) . A = 4 m, ω = 4 s⁻¹, Φ = (π/6) . At t = 0.25 : 4 × 0.25 + (π/6) = 1 + (π/6) ≈ 1.523 rad ≈ 87° . v = 4 × 4 cos 87° ≈ 16 × 0.052 ≈ 0.832 m/s . Applying x = A cos(ωt

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

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Forced Oscillations and Resonance

A particle in SHM has \( x = 4 \cos (2t - \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 2 s⁻¹, Φ = -(π/6) . At t = 0.5 : 2 × 0.5 - (π/6) = 1 - (π/6) ≈ 0.476 rad ≈ 27.3° . v = -2 × 4 sin (27.3°) ≈ -8 × 0.46 ≈ -3.68 m/s

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

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Forced Oscillations and Resonance

A particle in SHM has \( x = 3 \cos (2\pi t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)?

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 2π s⁻¹, Φ = (π/3) . At t = 0.5 : 2π × 0.5 + (π/3) = π + (π/3) = (4π/3) . v = -2π × 3 sin (4π/3) = -6π sin (180° - 60°) = -6π (-(√(3)/2)) ≈ 16.31 m/s . Applying x =

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

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( x = 4 \sin (3t) \) (in m). What is its speed at \( x = 2 \, \text{m} \)?

**SHM condition** is linear restoring force and inertia producing sinusoidal time dependence. Motions violating a = -ω² x, such as uniform circular motion, are periodic without oscillation about fixed point, highlighting classification criteria for NCERT. Velocity: v = ± ω √(A² - x²) . A = 4 m, ω = 3 s⁻¹, x = 2 m . v = 3 √(4² - 2²) = 3 √(16 - 4) = 3 √(12) = 3 × 2√(3) ≈ 10.39 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 10.39 m/s follows,

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

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Displacement, Velocity and Acceleration in SHM

A particle in SHM has \( x = 3 \cos (2t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 2 s⁻¹, Φ = (π/6) . At t = 0.5 : 2 × 0.5 + (π/6) = 1 + (π/6) = (π/3) + (π/6) = (π/2) . v = -2 × 3 sin ((π/2)) = -6 × 1 = -6 m/s . Applying x = A cos(ωt

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

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Displacement, Velocity and Acceleration in SHM

A particle in SHM has \( x = 6 \cos (2\pi t + \frac{\pi}{4}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)?

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 6 m, ω = 2π s⁻¹, Φ = (π/4) . At t = 0.25 : 2π × 0.25 + (π/4) = (π/2) + (π/4) = (3π/4) . v = -2π × 6 sin (3π/4) = -12π × (√(2)/2) ≈ -26.64 m/s . Applying x = A cos(ωt + φ), v =

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

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Energy Conservation and Friction

A 5kg block slides down a frictionless incline from 6m height. What is its speed at the bottom? (Take g\=10m/s2)

Potential energy mgh=5×10×6=300J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 11 m/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction

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Straight Line Motion - Practice Set

A 0.6kg pendulum bob completes a vertical circle of radius 1.2m. What is the speed at the top? (Take g\=10m/s2)

At top, minimum speed vC=gL=10×1.2=12≈3.46m/s. 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 3 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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Acceleration and Velocity Analysis

A ball is thrown horizontally at 11m/s from a height of 19.6m. What is its speed on hitting the ground? (Take g\=9.8m/s2

Vertical velocity: vy=2gh=2×9.8×19.6=384.16≈19.6m/s. 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 20 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

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Straight Line Motion - Mixed Concepts

A ball is thrown horizontally at 8m/s from a height of 12.5m. What is its speed on hitting the ground? (Take g\=10m/s2)

Vertical velocity: vy=2gh=2×10×12.5=250≈15.81m/s. 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 15 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 - Mixed Concepts

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