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#velocity calculation

12 public questions tagged with this topic.

Forced Oscillations and Resonance

A body oscillates with SHM according to \( x = 4 \cos (2\pi t + \frac{\pi}{6}) \) (in SI units). What is its velocity at

**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(t) = -ω A sin (ω t + Φ) . Here, A = 4 m, ω = 2π s⁻¹, Φ = (π/6) . At t = 0.5 s : ω t + Φ = 2π × 0.5 + (π/6) = π + (π/6) = (7π/6) . sin (7π/6) = sin (180° + 30°) = -sin 30° = -(1/2) . v = -2π

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

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Forced Oscillations and Resonance

A particle’s displacement is \( x = 5 \sin (2\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at \( t = 0.5 \, \te

**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. Velocity: v = ω A cos (ω t + Φ) . A = 5 m, ω = 2π s⁻¹, Φ = (π/6) . At t = 0.5 : 2π × 0.5 + (π/6) = π + (π/6) = (7π/6) . v = 2π × 5 cos (7π/6) = 10π cos (180° - 30°) = 10π (-(√(3)/2)) ≈ -27.14 m/s . Applying x = A cos(ωt + φ), v

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

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Displacement, Velocity and Acceleration in SHM

A particle’s displacement is \( x = 8 \sin (2\pi t - \frac{\pi}{3}) \) (in m). What is its velocity at \( t = 0.25 \, \t

**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 cos (ω t + Φ) . A = 8 m, ω = 2π s⁻¹, Φ = -(π/3) . At t = 0.25 : 2π × 0.25 - (π/3) = (π/2) - (π/3) = (π/6) . v = 2π × 8 cos (π/6) = 16π × (√(3)/2) ≈ 43.54 m/s . Applying x = A cos(ωt + φ), v =

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

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Motion Parameters and Calculations

A particle’s position is given by x\=2t2+3t and y\=t2−2t (in meters and seconds). What is the magnitude of its velocity

Velocity: vx=dxdt=4t+3,vy=dydt=2t−2. 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 10 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations

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Straight Line Motion - Mixed Concepts

A ball is thrown at 30m/s at 53∘. What is its vertical velocity at t\=2s? (Take g\=10m/s2,sin⁡53∘\=0.8)

Vertical velocity vy=v0sin⁡θ0−gt. 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 4 m/s as the result, so option C is correct.

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

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Motion in a Straight Line - Advanced Problems

A stone is dropped from rest. What is its velocity after falling 24.5m? (Take g\=10m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 25 m/s. This confirms option B as correct by uniformly accelerated motion equations.

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

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Distance, Displacement and Velocity-Time Relations

A stone is thrown upwards with a speed of 14m/s. What is its velocity after 1.5s? (Take g\=10m/s2)

Use v=v0+at. Here, v0=14m/s, a=−10m/s2, t=1.5s. 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 -1 m/s as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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Distance, Displacement and Velocity-Time Relations

A stone is dropped from rest. What is its velocity after falling 78.4m? (Take g\=9.8m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 40 m/s. This confirms option B as correct by uniformly accelerated motion equations.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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Distance, Displacement and Velocity-Time Relations

A ball is dropped from a height of 80m. What is its velocity just before hitting the ground? (Take g\=10m/s2)

For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 40 m/s. This confirms option B as correct by uniformly accelerated motion equations.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations

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