#circular motion

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

View Details

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

View Details

Damped Oscillations

A particle’s x-projection from circular motion is \( x = 6 \cos (3t) \) (in m). What is its maximum acceleration?

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Maximum acceleration: aₘₐₓ = ω² A . A = 6 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 6 = 9 × 6 = 54 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 54 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

View Details

Equations of SHM, Phase and Angular Frequency

Which statement best explains why uniform circular motion is not considered oscillatory despite being periodic?

**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°. Oscillatory motion requires to-and-fro movement about an equilibrium, while uniform circular motion involves continuous rotation without reversing direction. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It does not involve to-and-fro motion follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details

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

View Details

Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 7 \cos (3t) \) (in m). What is its maximum acceleration?

**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). Maximum acceleration: aₘₐₓ = ω² A . A = 7 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 7 = 9 × 7 = 63 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 63 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details

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

View Details

Displacement, Velocity and Acceleration in SHM

A particle’s x-projection from circular motion is \( x = 8 \cos (2\pi t) \) (in m). What is its maximum acceleration?

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Maximum acceleration: aₘₐₓ = ω² A . A = 8 m, ω = 2π s⁻¹ . aₘₐₓ = (2π)² × 8 ≈ 39.48 × 8 ≈ 315.84 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E =

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

View Details

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

View Details

Displacement, Velocity and Acceleration in SHM

The projection of a particle in circular motion on the x-axis is \( x = 3 \cos (2\pi t) \) (in m). What is the radius of

**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. For SHM as projection of circular motion, radius = amplitude. Here, x = A cos (ω t) , so A = 3 m . Radius = 3 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.0 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

View Details

PHYSICS

A 0.3 kg stone is whirled at 50 rev/min in a circle of radius 2 m with a 5 N tangential force. What is the net force at

Given: A 0.3 kg stone is whirled at 50 rev/min in a circle of radius 2 m with a 5 N tangential force. What is the net force at that instant? These values define the system as per NCERT data. Formula: Centripetal force: F_c = m omega² r, omega = 50 × 2π/60 = 5π/3 rad/s. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: omega² = (5π/3)² = 25π²/9 . F_c = 0.3 × 25π²/9 × 2 approx 0.3 × 54.74 approx 16.42 N . Tangential force = 5 N . Net force F = sqrt(F_c)² + (F_t)² = sqrt(16.42)² + (5)² approx sqrt269.62 + 25 approx 17.16 N . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

View Details

PHYSICS

A 0.3 kg stone is whirled in a horizontal circle of radius 1.6 m at 45 rev/min . What is the tension? (Take g = 10 m/s²

Given: A 0.3 kg stone is whirled in a horizontal circle of radius 1.6 m at 45 rev/min . What is the tension? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Angular speed: omega = 45 × 2π/60 = 3π/2 rad/s. This is standard NCERT relation. Substitution & Calculation: omega² = (3π/2)² = 9π²/4 approx 22.21 . Tension: T = m omega² r = 0.3 × 22.21 × 1.6 . T approx 0.3 × 22.21 × 1.6 approx 10.66 N . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details