#velocity

Forced Oscillations and Resonance

A particle’s motion is \( x = 4 \sin (3t - \frac{\pi}{3}) \) (in m). What is its velocity at \( t = \frac{\pi}{6} \, \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 = 4 m, ω = 3 s⁻¹, Φ = -(π/3) . At t = (π/6) : 3 × (π/6) - (π/3) = (π/2) - (π/3) = (π/6) . v = 3 × 4 cos (π/6) = 12 × (√(3)/2) = 6√(3) ≈ 10.39 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt

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

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Damped Oscillations

A mass oscillates with \( v = -8 \cos (2t) \) (in m/s). What is its displacement function?

**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. Velocity: v = -ω A sin (ω t) , but given v = -8 cos (2t) . ω = 2 s⁻¹, vₘₐₓ = ω A = 8 ⇒ A = (8/2) = 4 m . Since v = -A ω sin (ω t) , adjust phase: x = 4 sin (2t) . Applying x = A cos(ωt + φ), v = -ωA

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Damped Oscillations

A particle’s displacement is \( x = 3 \cos (2\pi t - \frac{\pi}{4}) \) (in m). What is its velocity at \( t = 0.25 \, \t

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 2π s⁻¹, Φ = -(π/4) . At t = 0.25 : 2π × 0.25 - (π/4) = (π/2) - (π/4) = (π/4) . v = -2π × 3 sin (π/4) = -6π × (√(2)/2) ≈ -13.32 m/s . Applying x = A cos(ωt + φ),

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Energy in SHM - Kinetic, Potential and Total

A particle’s displacement is \( x = 4 \cos (3\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at \( t = 0 \, \text

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 3π s⁻¹, Φ = (π/6) . At t = 0 : v = -3π × 4 sin (π/6) = -12π × 0.5 ≈ -18.84 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ),

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

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Periodic Motion and SHM Basic Concepts

A particle’s displacement is \( x = 3 \cos (4\pi t + \frac{\pi}{3}) \) (in m). What is its velocity at \( t = 0 \, \text

**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 4π s⁻¹, Φ = (π/3) . At t = 0 : v = -4π × 3 sin (π/3) = -12π × (√(3)/2) ≈ -32.58 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and

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

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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

In SHM, why does the particle’s velocity lead its displacement by \( \pi/2 \) radians?

**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. Displacement ( x = A cos (ω t + Φ) ) and velocity ( v = -ω A sin (ω t + Φ) ) differ by π/2 radians because the cosine and sine functions are shifted by this phase, reflecting their derivative relationship. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² =

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

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Displacement, Velocity and Acceleration in SHM

A particle’s displacement is \( x = 7 \sin (2t + \frac{\pi}{4}) \) (in m). What is its velocity at \( t = 0 \, \text{s}

**Velocity and acceleration in SHM** follow from differentiation, showing 90° phase lead of v over x and 180° for a over x. At mean position x=0, a=0, v=±ωA maximum; at extremes x=±A, v=0, a=∓ω²A maximum magnitude, illustrating energy conversion. Velocity: v = ω A cos (ω t + Φ) . A = 7 m, ω = 2 s⁻¹, Φ = (π/4) . At t = 0 : v = 2 × 7 cos (π/4) = 14 × (√(2)/2) = 7√(2) ≈ 9.9 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² =

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

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PHYSICS

Check the dimensional consistency of v = sqrtP/rho, where v is velocity, P is pressure, and rho is density.

LHS: [v] = [L T^{-1] . RHS: [P / rho] = [M L^{-1 T^{-2] / [M L^{-3] = [L² T^{-2] . sqrt[L² T^{-2] = [L T^{-1], consistent. This follows from NCERT principle where the relation explains the outcome clearly for students in simple steps.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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Surface Tension, Capillarity and Excess Pressure

Which statement is incorrect about Bernoulli’s principle?

Bernoulli’s principle assumes non-viscous flow, not high viscosity, as viscosity causes energy loss, violating the energy conservation assumption. The incorrect statement is that it applies to highly viscous fluids. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It applies to highly viscous fluids. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A 6kg particle moves with velocity v\=3j^m/s at r\=−4i^m. What is the magnitude of its angular momentum about the origin

L = r×p = |i^j^k^−400030| = k^((−4)×3−0×0) = −12k^kg m2/s. Magnitude = 12kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 12 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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Moment of Inertia and Radius of Gyration

A 4kg particle moves with velocity v\=3i^+5j^m/s at r\=−2i^m. What is the z-component of its angular momentum?

L = r×p = |i^j^k^−200350| = k^((−2)×5−0×3) = −10k^kg m2/s. Z-component = −10kg m2/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -10 kg m²/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

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