Skip to content

#maximum velocity

5 public questions tagged with this topic.

A particle in SHM has an amplitude of \( 8 \, \text{cm} \) and a period of \( 0.4 \, \text{s} \). What is its maximum ve

**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°. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/0.4) = 15.7 rad/s . A = 0.08 m . vₘₐₓ = 0.08 × 15.7 = 1.256 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.256 m/s follows, reflecting SHM dependence on

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

A particle in SHM has an amplitude of \( 9 \, \text{cm} \) and a period of \( 0.8 \, \text{s} \). What is its maximum ve

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. Maximum velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/0.8) = 7.85 rad/s . A = 0.09 m . vₘₐₓ = 0.09 × 7.85 ≈ 0.7065 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.7065 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

A particle in SHM has an amplitude of \( 6 \, \text{cm} \) and a frequency of \( 3 \, \text{Hz} \). What is its maximum

**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. Maximum velocity: vₘₐₓ = A ω . ω = 2π v = 2 × 3.14 × 3 = 18.84 rad/s . A = 0.06 m . vₘₐₓ = 0.06 × 18.84 = 1.1304 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.1304 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

A particle in SHM has an amplitude of \( 12 \, \text{cm} \) and a period of \( 1.2 \, \text{s} \). What is its maximum v

**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 velocity: vₘₐₓ = A ω . ω = (2π/T) = (2 × 3.14/1.2) ≈ 5.23 rad/s . A = 0.12 m . vₘₐₓ = 0.12 × 5.23 ≈ 0.628 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.628 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

A particle in SHM has an amplitude of \( 10 \, \text{cm} \) and a frequency of \( 2.5 \, \text{Hz} \). What is its maxim

**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. Maximum velocity: vₘₐₓ = A ω . ω = 2π v = 2 × 3.14 × 2.5 = 15.7 rad/s . A = 0.1 m . vₘₐₓ = 0.1 × 15.7 = 1.57 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.57 m/s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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