IGNOU BPHCT-131 Solved Question Paper PDF Download

IGNOU BPHCT-131 Previous Year Solved Question Paper in Hindi

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Q1. (a) Determine the angle between any two vectors p and q of non-zero magnitude given that: |p+q| = |p-q| (b) State the order and degree of the following differential equation: y”’ + √(y”) = x (c) Astronauts are put in a rotating circular chamber to be trained in how to withstand high accelerations. What is the acceleration experienced by an astronaut in a chamber in terms of g if it rotates with a constant angular speed of 2.0 rad s⁻¹? Given that in the chamber astronauts travel in a circle of radius of 10 m. (d) A lift of mass 3000 kg moves 200 m upward at a constant speed in 25.0 s. At what average rate does the force due to the cable do work on the lift? Take g=10ms⁻². (e) The rotational kinetic energy of a particle in circular motion is 100 J. It moves once in the circle in 22.0 s. What is the rotational inertia of the particle? (f) Using the law of equal areas, explain why an object would move slower when it is farther away from the centre of force. (g) The displacement of an object executing simple harmonic motion is given by: x = 0.05cos(4πt + 0.0625)m. Determine: (i) maximum velocity (ii) maximum acceleration (h) What is relaxation time for a damped oscillation?

Ans. (a) Given the condition: | p + q | = | p – q | Squaring both sides of the equation: | p + q |² = | p – q |² Using the property | A |² = A ⋅ A : ( p + q ) ⋅ ( p + q ) = ( p – q ) ⋅ ( p – q ) p ⋅ p + p ⋅ q + q ⋅ p + q ⋅ q = p ⋅ p – p ⋅ q – q ⋅ p + q ⋅ q p² + 2( p ⋅ q ) + q² = p² – 2( p ⋅ q ) + q² 4( p ⋅ q ) = 0 p ⋅ q = 0 If θ is the angle between vectors p and q , then p ⋅ q = pq cosθ. pq cosθ = 0 Since p and q are non-zero vectors (p ≠ 0, q ≠ 0), it must be that cosθ = 0. Therefore, θ = 90° or π/2 radians. The two vectors are perpendicular to each other.

(b) The given differential equation is: y”’ + √(y”) = x To determine the degree, we must first make the equation a polynomial in its derivatives. We isolate the radical term: y”’ – x = -√(y”) Squaring both sides: (y”’ – x)² = (-√(y”))² (y”’)² – 2xy”’ + x² = y” The order of a differential equation is the order of the highest derivative present. Here, the highest derivative is y”’. Thus, the order is 3 . The degree of a differential equation is the power of the highest order derivative after the equation has been rationalized. Here, the power of the highest derivative (y”’) is 2. Thus, the degree is 2 .

(c) Given: Angular speed, ω = 2.0 rad s⁻¹, Radius, r = 10 m. In uniform circular motion, the centripetal acceleration of the particle is given by: a = ω²r a = (2.0 rad s⁻¹)² × 10 m = 4.0 s⁻² × 10 m = 40 m/s² To express the acceleration in terms of g (where g ≈ 10 m/s²), we divide the acceleration ‘a’ by ‘g’: a/g = (40 m/s²) / (10 m/s²) = 4 Hence, the acceleration is a = 4g .

(d) Given: Mass, m = 3000 kg, Distance, d = 200 m, Time, t = 25.0 s, g = 10 m/s². Since the lift moves upward at a constant speed, the net force on it is zero. The force exerted by the cable (F) is equal to the weight of the lift (mg). F = mg = 3000 kg × 10 m/s² = 30000 N The work done (W) by this force is: W = F × d = 30000 N × 200 m = 6,000,000 J The average rate of doing work (Power, P) is: P = W / t = 6,000,000 J / 25.0 s = 240,000 W or 2.4 × 10⁵ W .

(e) Given: Rotational kinetic energy, KE_rot = 100 J, Time for one revolution (Period), T = 22.0 s. The angular speed (ω) of the particle is: ω = 2π / T = 2π / 22.0 s = π / 11 rad/s The formula for rotational kinetic energy is: KE_rot = (1/2)Iω² Where I is the rotational inertia. 100 J = (1/2) × I × (π / 11 s⁻¹)² 100 = (1/2) × I × (π²/121) I = (100 × 2 × 121) / π² = 24200 / π² Using π² ≈ (3.14159)² ≈ 9.87: I ≈ 24200 / 9.87 ≈ 2451.9 kg m² .

(f) The law of equal areas (Kepler’s Second Law) states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This means the areal velocity (dA/dt) is constant. The areal velocity is related to the angular momentum (L) and mass (m) of the object by dA/dt = L / (2m). Since angular momentum is conserved under a central force, L and m are constant, hence dA/dt is constant. The area swept in a small time interval dt is dA ≈ (1/2)r(v⊥dt), where v⊥ is the component of velocity perpendicular to the radius vector. Thus, dA/dt = (1/2)rv⊥ = constant. From this relationship, if the object is farther from the centre of force (r is large), the perpendicular component of its velocity, v⊥, must be smaller to keep the product constant. Therefore, an object moves slower when it is farther away from the centre of force.

(g) The given displacement equation is: x = 0.05cos(4πt + 0.0625) m. Comparing this with the standard SHM equation x = Acos(ωt + φ), we identify: Amplitude, A = 0.05 m Angular frequency, ω = 4π rad/s (i) Maximum velocity (v_max) is given by: v_max = Aω = 0.05 m × 4π rad/s = 0.2π m/s ≈ 0.628 m/s . (ii) Maximum acceleration (a_max) is given by: a_max = Aω² = 0.05 m × (4π rad/s)² = 0.05 × 16π² m/s² = 0.8π² m/s² ≈ 7.89 m/s² .

(h) In a damped oscillation, the relaxation time (τ) is the time taken for the amplitude of the oscillation to decay to 1/e (approximately 37%) of its initial value. It is a measure of the rate of damping. If the amplitude is given by the equation A(t) = A₀e⁻ᵇᵗ/(²ᵐ), the relaxation time is τ = 2m/b, where ‘b’ is the damping coefficient and ‘m’ is the mass. A shorter relaxation time indicates heavier damping.

Q2. (a) The position vector of a particle as a function of time is given by: r(t) = 5cos(4t)î + 5sin(4t)ĵ + 10k. Determine its velocity and acceleration. Show that both its speed and the magnitude of its acceleration are constant. (b) Show that the following ODE is exact and solve it: xy’ + 2x + y = 0 (c) In an LCR circuit, an inductance L, a resistance R and a capacitance C are connected in series. The variation of charge ‘q’ flowing through with time ‘t’ in the circuit is given by the differential equation: L(d²q/dt²) + R(dq/dt) + q/C = 0. Solve this equation to determine ‘q’ as a function of time ‘t’.

Ans. (a) The given position vector is: r (t) = 5cos(4t)î + 5sin(4t)ĵ + 10k The velocity (v) is the first derivative of the position vector with respect to time: v (t) = d r /dt = d/dt [5cos(4t)î + 5sin(4t)ĵ + 10k] v (t) = 5(-4sin(4t))î + 5(4cos(4t))ĵ + 0k v (t) = -20sin(4t)î + 20cos(4t)ĵ The acceleration (a) is the first derivative of the velocity vector with respect to time: a (t) = d v /dt = d/dt [-20sin(4t)î + 20cos(4t)ĵ] a (t) = -20(4cos(4t))î + 20(-4sin(4t))ĵ a (t) = -80cos(4t)î – 80sin(4t)ĵ Now, we find the speed (magnitude of velocity) and the magnitude of acceleration. Speed (|v|) : | v (t)| = √[(-20sin(4t))² + (20cos(4t))²] | v (t)| = √[400sin²(4t) + 400cos²(4t)] | v (t)| = √[400(sin²(4t) + cos²(4t))] = √400 = 20 m/s Since the speed is 20 m/s, a constant value, it does not depend on time. Thus, the speed is constant . Magnitude of acceleration (|a|) : | a (t)| = √[(-80cos(4t))² + (-80sin(4t))²] | a (t)| = √[6400cos²(4t) + 6400sin²(4t)] | a (t)| = √[6400(cos²(4t) + sin²(4t))] = √6400 = 80 m/s² Since the magnitude of acceleration is 80 m/s², a constant value, it also does not depend on time. Thus, the magnitude of acceleration is constant .

(b) The given differential equation is: xy’ + 2x + y = 0 Using y’ = dy/dx and rearranging: x(dy/dx) + (2x + y) = 0 Multiplying by dx: x dy + (2x + y) dx = 0 Arranging in the form M(x, y)dx + N(x, y)dy = 0: (y + 2x)dx + x dy = 0 Here, M(x, y) = y + 2x and N(x, y) = x. The condition for exactness is ∂M/∂y = ∂N/∂x. ∂M/∂y = ∂/∂y (y + 2x) = 1 ∂N/∂x = ∂/∂x (x) = 1 Since ∂M/∂y = ∂N/∂x, the given ODE is exact . Solution : The solution of an exact equation is given by ∫M dx (treating y as constant) + ∫(terms in N not containing x) dy = C. ∫(y + 2x) dx + ∫(0) dy = C xy + x² = C Thus, the solution to the given differential equation is xy + x² = C , where C is the constant of integration.

(c) The given differential equation for an LCR circuit is: L(d²q/dt²) + R(dq/dt) + q/C = 0 This is a second-order linear homogeneous differential equation with constant coefficients. To solve it, we form the characteristic (or auxiliary) equation by assuming a solution of the form q = e^(αt): Lα² + Rα + 1/C = 0 The roots of this quadratic equation are: α = [-R ± √(R² – 4L/C)] / (2L) The nature of the solution depends on the sign of the discriminant, D = R² – 4L/C.

• Case 1: Overdamped (R² > 4L/C) In this case, the discriminant is positive, and there are two real, distinct roots for α, let’s call them α₁ and α₂. α₁, α₂ = -R/(2L) ± √((R/2L)² – 1/(LC)) The general form of the solution is: q(t) = A₁e^(α₁t) + A₂e^(α₂t) where A₁ and A₂ are constants determined by initial conditions. The charge decays exponentially to zero without oscillation.
• Case 2: Critically Damped (R² = 4L/C) In this case, the discriminant is zero, and there is one real, repeated root for α, which is α = -R/(2L). The general form of the solution is: q(t) = (A₁ + A₂t)e^(-Rt/2L) This represents the fastest possible decay without any oscillation.
• Case 3: Underdamped (R² < 4L/C) In this case, the discriminant is negative, and the roots are complex conjugates. α = -R/(2L) ± i√((1/LC) – (R/2L)²) This can be written as α = -γ ± iω’, where γ = R/2L (the damping constant) and ω’ = √((1/LC) – (R/2L)²) is the damped angular frequency. The general form of the solution is: q(t) = e^(-γt) [A₁cos(ω’t) + A₂sin(ω’t)] Or, equivalently, q(t) = A e^(-γt) cos(ω’t + φ), where A and φ are constants. The charge oscillates with a decaying amplitude over time.

Q3. (a) The mass of an aircraft is 50000 kg. It is flying in a straight line at a constant speed of 1000 km h⁻¹. The weight of the aircraft equals the lift force. The pilot increases the thrust of the engine to 90000 N. Suppose the air resistance force equals the engine thrust in 25 s. What is the increased constant speed of the aircraft at that instant? What is the increase in its speed? (b) Calculate the height of a geosynchronous satellite above the surface of the Earth. Given that: G = 6.67×10⁻¹¹ Nm²kg⁻², Mass of Earth Mₑ = 6.0×10²⁴ kg and Radius of Earth = 6.4×10⁶ m. (c) The mass of a two-stage rocket launched in free space is 1000 kg at some instant of time. When 600 kg of fuel of the first stage burns in it, the rocket ejects a stream of gas at a relative velocity of 500 ms⁻¹. What is the rocket’s velocity after the first stage is ejected?

Ans. (a) The problem is slightly ambiguous about how the air resistance changes. We will make a reasonable assumption to solve it. Given: Mass, m = 50000 kg Initial speed, vᵢ = 1000 km/h = 1000 × (1000 m / 3600 s) ≈ 277.8 m/s New thrust, T_new = 90000 N Time, t = 25 s After 25 s, at the new constant speed, the air resistance R_f equals the new thrust, so R_f = 90000 N. Initially, at constant speed, the initial thrust T_i was equal to the initial air resistance R_i. The value of R_i is unknown. Assumption: The net force causing acceleration changes over the 25s period (from F_net,initial = 90000 – R_i to F_net,final = 0). A simple model is to assume a constant average acceleration, which implies a constant average net force. Let’s assume the average net force is half of the initial net force, F_avg = (90000 – R_i)/2. As R_i is unknown, we make a further simplification by assuming the average accelerating force can be approximated as half the maximum possible net force, say (90000 N)/2 = 45000 N. This is a significant assumption made to proceed. Average acceleration, a_avg = F_avg / m = 45000 N / 50000 kg = 0.9 m/s² We can now find the final speed (v_f) using the equation of motion: v_f = v_i + a_avg × t v_f = 277.8 m/s + (0.9 m/s² × 25 s) v_f = 277.8 m/s + 22.5 m/s = 300.3 m/s In km/h, this is: 300.3 × (3600/1000) ≈ 1081 km/h. Increase in speed : Δv = v_f – v_i = 300.3 m/s – 277.8 m/s = 22.5 m/s .

(b) A geosynchronous satellite orbits the Earth with a period of 24 hours, matching the Earth’s rotational period. Orbital period, T = 24 hours = 24 × 3600 s = 86400 s For an orbit of radius ‘r’ (from the center of the Earth), the gravitational force provides the necessary centripetal force: F_g = F_c GMₑm / r² = mω²r where m is the satellite’s mass and ω is the angular velocity. GMₑ / r³ = ω² Since ω = 2π / T, we can write: GMₑ / r³ = (2π / T)² = 4π² / T² r³ = (GMₑT²) / (4π²) Plugging in the given values: G = 6.67 × 10⁻¹¹ Nm²kg⁻² Mₑ = 6.0 × 10²⁴ kg T = 86400 s r³ = [(6.67 × 10⁻¹¹) × (6.0 × 10²⁴) × (86400)²] / [4π²] r³ = [4.002 × 10¹⁴ × 7.465 × 10⁹] / [39.48] r³ ≈ (2.988 × 10²⁴) / 39.48 ≈ 7.568 × 10²² m³ r = (7.568 × 10²²)¹/³ m ≈ 4.23 × 10⁷ m This is the distance from the Earth’s center. The height (h) above the Earth’s surface is: h = r – Rₑ where Rₑ is the radius of the Earth = 6.4 × 10⁶ m = 0.64 × 10⁷ m. h = (4.23 × 10⁷ m) – (0.64 × 10⁷ m) h = 3.59 × 10⁷ m = 35900 km So, the height of a geosynchronous satellite is approximately 35,900 km.

(c) We can use the Tsiolkovsky rocket equation to find the change in velocity: Δv = v_rel * ln(m_i / m_f) Where: Δv = change in rocket’s velocity v_rel = relative velocity of the exhaust gas = 500 m/s m_i = initial mass of the rocket = 1000 kg m_f = final mass of the rocket (after fuel is burnt) Mass of fuel burnt = 600 kg. Final mass, m_f = m_i – (fuel burnt) = 1000 kg – 600 kg = 400 kg. Substituting the values into the equation: Δv = 500 × ln(1000 / 400) Δv = 500 × ln(2.5) Using ln(2.5) ≈ 0.9163: Δv = 500 × 0.9163 ≈ 458.15 m/s Since the rocket is “launched in free space”, we can assume it starts from rest at that instant (initial velocity = 0). Therefore, the rocket’s velocity after the first stage is ejected will be equal to Δv. The rocket’s velocity ≈ 458 m/s .

Q4. (a) (i) A bicycle travels 50 m along a circular track of radius 5 m. What is the angular displacement in radians of the bicycle from its starting position? (ii) The Earth orbits the sun in 365.25 days in a nearly circular orbit. What is the average angular speed of a particle on Earth’s surface as it orbits the sun? Take the direction of Earth’s rotation to be +ve direction of the angular displacement. (b) State the law of conservation of angular momentum. A merry-go-round possessing rotational inertia 5000 kgm² is mounted on a frictionless vertical axle and is initially rotating at an angular speed of 1 revolution per minute. A girl jumps on to the platform in the radial direction. If the rotational speed of the merry-go-round reduces to 0.8 r.p.m., calculate the girl’s rotational inertia. (c) A billiard ball of mass ‘m’ hits another billiard ball of equal mass at rest in an elastic collision and moves along a straight line at an angle of θ from its original direction of motion. At what angle with each other do the target ball and the projectile move after collision?

Ans. (a) (i) Given: Distance traveled (arc length), s = 50 m Radius, r = 5 m The angular displacement (θ) in radians is given by the ratio of the arc length to the radius: θ = s / r θ = 50 m / 5 m = 10 radians The angular displacement of the bicycle is 10 radians.

(ii) The period of Earth’s orbit around the sun is: T = 365.25 days Converting this to seconds: T = 365.25 days × 24 hours/day × 3600 seconds/hour ≈ 3.156 × 10⁷ s The angular displacement for one full orbit is Δθ = 2π radians. The average angular speed (ω) is: ω = Δθ / T = 2π / T ω = 2π radians / (3.156 × 10⁷ s) ω ≈ 1.99 × 10⁻⁷ rad/s Since the direction of rotation is taken as positive, the angular speed is positive.

(b) Law of Conservation of Angular Momentum: This law states that if the net external torque acting on a system is zero, the total angular momentum of the system remains constant (or conserved). Mathematically, if τ_ext = 0, then d L /dt = 0, which implies L = constant. Numerical Problem: The system is the merry-go-round + girl. Since the girl jumps radially, she brings no initial angular momentum, and there is no external torque. Thus, the angular momentum of the system is conserved. L_initial = L_final I_initial ω_initial = I_final ω_final Given: Rotational inertia of merry-go-round, I_mgr = 5000 kg m² Initial angular speed, ω_i = 1 rpm (revolutions per minute) Final angular speed, ω_f = 0.8 rpm Initially, only the merry-go-round is rotating: I_initial = I_mgr = 5000 kg m² Finally, the girl and merry-go-round rotate together: I_final = I_mgr + I_girl Applying the conservation law: I_mgr ω_i = (I_mgr + I_girl) ω_f 5000 kg m² × 1 rpm = (5000 kg m² + I_girl) × 0.8 rpm (We can keep the units in rpm as they cancel from both sides) 5000 = (5000 + I_girl) × 0.8 5000 / 0.8 = 5000 + I_girl 6250 = 5000 + I_girl I_girl = 6250 – 5000 = 1250 kg m² Thus, the girl’s rotational inertia is 1250 kg m².

(c) This is a special case of a two-dimensional elastic collision between two objects of equal mass, with one initially at rest. Let the projectile ball have mass m and initial velocity u₁, and the target ball have mass m and initial velocity u₂ = 0. After the collision, the projectile ball has velocity v₁ and makes an angle θ with the initial direction. The target ball has velocity v₂ and makes an angle φ with the initial direction. From conservation of momentum : x-axis: mu₁ = mv₁cosθ + mv₂cosφ => u₁ = v₁cosθ + v₂cosφ (1) y-axis: 0 = mv₁sinθ – mv₂sinφ => v₁sinθ = v₂sinφ (2) From conservation of kinetic energy (elastic collision): (1/2)mu₁² = (1/2)mv₁² + (1/2)mv₂² => u₁² = v₁² + v₂² (3) It is a well-known result that when a moving object elastically collides with another object of the same mass at rest (and the collision is not head-on), the two objects move off at right angles (90°) to each other after the collision. Proof: The momentum conservation equations can be written in vector form as: m u₁ = m v₁ + m v₂ => u₁ = v₁ + v₂ Taking the dot product of this vector equation with itself: u₁ ⋅ u₁ = ( v₁ + v₂ ) ⋅ ( v₁ + v₂ ) u₁² = v₁ ⋅ v₁ + 2( v₁ ⋅ v₂ ) + v₂ ⋅ v₂ u₁² = v₁² + v₂² + 2( v₁ ⋅ v₂ ) (4) Substituting the energy conservation equation (3) into equation (4): v₁² + v₂² = v₁² + v₂² + 2( v₁ ⋅ v₂ ) 0 = 2( v₁ ⋅ v₂ ) v₁ ⋅ v₂ = 0 The dot product of two non-zero vectors is zero if and only if they are perpendicular to each other. Therefore, the two balls move at 90° to each other after the collision. The angle between the target and the projectile is 90° or π/2 .

Q5. (a) The amplitude of vibration of a damped spring mass system decreases from 10 cm to 2.5 cm in 200 s. If this oscillator completes 50 oscillations in this time, compare the periods with and without damping. (b) Two collinear harmonic oscillations are represented by: x₁(t) = 4sin(20πt + π/3) cm, x₂(t) = 3sin(20πt + π/6) cm. Calculate the amplitude, phase constant and period of resultant oscillation obtained on superposing these two collinear oscillations. (c) (i) How does a pulse differ from a wave? (ii) A progressive transverse wave is described by: y(x, t) = 0.01sin(256t – 63x) m. Determine the direction of propagation of the wave and calculate the amplitude, wavelength, frequency and velocity.

Ans. (a) For a damped oscillator, the amplitude decays exponentially with time: A(t) = A₀e⁻γᵗ where A₀ is the initial amplitude, A(t) is the amplitude at time t, and γ is the damping factor. Given: A₀ = 10 cm, A(200s) = 2.5 cm, t = 200 s. 2.5 = 10 e⁻²⁰⁰γ 0.25 = e⁻²⁰⁰γ Taking the natural logarithm of both sides: ln(0.25) = -200γ -ln(4) = -200γ => γ = ln(4) / 200 ≈ 1.386 / 200 ≈ 0.00693 s⁻¹. The oscillator completes 50 oscillations in 200 s. Therefore, the damped period (T_d) is: T_d = 200 s / 50 = 4.0 s The damped angular frequency (ω_d) is: ω_d = 2π / T_d = 2π / 4 = π/2 rad/s ≈ 1.571 rad/s. The relationship between the natural angular frequency (ω₀, without damping) and the damped angular frequency (ω_d) is: ω_d² = ω₀² – γ² ω₀² = ω_d² + γ² ω₀² = (π/2)² + (0.00693)² ≈ (1.571)² + 0.000048 ≈ 2.468 + 0.000048 = 2.468048 ω₀ = √2.468048 ≈ 1.571002 rad/s The undamped period (T₀) is: T₀ = 2π / ω₀ = 2π / 1.571002 ≈ 3.999 s Comparison: The damped period is T_d = 4.0 s, while the undamped period is T₀ ≈ 3.999 s. The two periods are nearly equal (T_d ≈ T₀). This is because the damping is very weak (the value of γ is much smaller than ω_d), so it does not significantly affect the frequency of oscillation.

(b) The two given collinear harmonic oscillations are: x₁(t) = 4sin(20πt + π/3) x₂(t) = 3sin(20πt + π/6) By the principle of superposition, the resultant displacement is x(t) = x₁(t) + x₂(t). The resultant oscillation will have the form x(t) = A_R sin(20πt + δ), where A_R is the resultant amplitude and δ is the resultant phase constant. We use the vector (phasor) addition method: A₁ = 4, φ₁ = π/3 A₂ = 3, φ₂ = π/6 Resultant Amplitude (A_R) : A_R² = A₁² + A₂² + 2A₁A₂cos(φ₁ – φ₂) The phase difference is φ₁ – φ₂ = π/3 – π/6 = π/6. A_R² = 4² + 3² + 2(4)(3)cos(π/6) = 16 + 9 + 24(√3/2) = 25 + 12√3 A_R² ≈ 25 + 12(1.732) = 25 + 20.784 = 45.784 A_R = √45.784 ≈ 6.77 cm Phase Constant (δ) : tan(δ) = (A₁sinφ₁ + A₂sinφ₂) / (A₁cosφ₁ + A₂cosφ₂) sin(π/3) = √3/2, cos(π/3) = 1/2 sin(π/6) = 1/2, cos(π/6) = √3/2 tan(δ) = [4(√3/2) + 3(1/2)] / [4(1/2) + 3(√3/2)] = (2√3 + 1.5) / (2 + 1.5√3) tan(δ) ≈ (2×1.732 + 1.5) / (2 + 1.5×1.732) = (3.464 + 1.5) / (2 + 2.598) = 4.964 / 4.598 ≈ 1.08 δ = arctan(1.08) ≈ 0.824 radians (or 47.2°) Period (T) : The angular frequency ω = 20π rad/s is the same for both oscillations. Therefore, the resultant oscillation will also have an angular frequency of 20π rad/s. T = 2π / ω = 2π / (20π) = 0.1 s .

(c) (i) Difference between a pulse and a wave:

• Pulse: A pulse is a single, non-periodic disturbance that travels through a medium. It has a definite beginning and end and lasts for a short duration. Example: A single flick of one end of a rope.
• Wave: A wave (or wave train) is a continuous, periodic disturbance that transfers energy and momentum through a medium. It can be thought of as a succession of pulses repeated at regular intervals. Example: A sinusoidal wave.

The key differences lie in their

duration

and

periodicity

.

(ii) The given progressive wave equation is: y(x, t) = 0.01sin(256t – 63x) m We compare this with the standard form y(x, t) = Asin(ωt – kx):

• Direction of Propagation: Since the sign between the ‘t’ and ‘x’ terms is negative (-), the wave is propagating in the positive x-direction .
• Amplitude (A): A = 0.01 m .
• Angular Frequency (ω): ω = 256 rad/s. The Frequency (f) = ω / 2π = 256 / (2π) ≈ 40.74 Hz .
• Wave Number (k): k = 63 rad/m. The Wavelength (λ) = 2π / k = 2π / 63 ≈ 0.0997 m .
• Velocity (v): The wave velocity is given by v = ω / k. v = 256 rad/s / 63 rad/m ≈ 4.06 m/s .