Physics 1st Ch 09 Wave

Physics – First Paper Chapter 9: Wave Curated Questions and Complete Model Solutions (HSC English Version) Chapter 9: Wave B. Creative Questions 1. The figure below illustrates a transverse ripple wave produced on the surface of water after throwing a small stone. The frequency of the wave is n = 40 Hz. The horizontal distance between crest A and trough D is 4.0 m. Displacement of particles A (Crest) B C (Trough) D E F 4 m n = 40 Hz Direction of propagation (a) What is a wavefront? (b) Explain the principle of superposition of waves. (c) Using the data given in the figure, calculate the wavelength and wave speed of the water wave. (d) Contrast the wave speed with the maximum instantaneous speed of the water particles if the amplitude of oscillation is 5 cm. Answer: (a) Wavefront: A wavefront is the continuous locus of all points in a medium that vibrate in the exact same phase. (b) Principle of Superposition: When two or more waves travel through a medium simul- taneously and overlap, the resultant displacement at any point at any instant is the vector sum of the individual displacements produced by each wave independently:⃗ y =⃗y1 +⃗y2 + · · · +⃗yn (c) Wavelength and wave speed: F

Given frequency n = 40 Hz, the wave speed is: v = nλ = 40 × 5.333 = 213.33 m/s (If the distance of 4 m is measured between consecutive crest A and trough C, λ 2 = 4 m =⇒ λ = 8 m, giving v = 40 × 8 = 320 m/s.) (d) Comparison of wave speed with particle speed: The wave speed is the uniform speed at which energy and phase propagate through the medium: vwave = 213.33 m/s The water particles execute simple harmonic motion perpendicular to the direction of wave travel. The maximum speed of a particle occurs as it passes through the mean position: vp, max = ωA = 2πnA Given amplitude A = 5 cm = 0.05 m and frequency n = 40 Hz: vp, max = 2 × 3.1416 × 40 × 0.05 = 4π ≈12.57 m/s Taking the ratio: vwave vp, max = 213.33 12.57 ≈17.0 The wave travels forward at a constant speed of 213.33 m/s, whereas the medium particles oscillate about their equilibrium positions with a varying speed that reaches a maximum of only 12.57 m/s. 2. The equation of a progressive wave traveling along the positive x-direction through a medium is: y = 5 sin(200πt −1.57x) where x and y are in meters and t is in seconds. (a) What is a progressive (traveling) wave? (b) What is meant by the intensity of sound? (c) Determine

1. Amplitude: A = 5 m 2. Frequency: ω = 200π =⇒2πf = 200π =⇒f = 100 Hz 3. Wavelength: k = 1.57 =⇒2π λ = 1.57 =⇒λ = 2 × 3.1416 1.57 ≈4.0 m 4. Wave velocity: v = fλ = 100 × 4.0 = 400 m/s  or v = ω k = 200π 1.57 ≈400 m/s  (d) Mathematical analysis of particle acceleration: The wave velocity v = 400 m/s repre- sents the constant speed of energy propagation through the medium: vwave = dx dt = constant However, individual particles of the medium oscillate about their fixed equilibrium positions with simple harmonic motion. The instantaneous particle displacement is: y(t) = 5 sin(200πt −1.57x) The particle velocity is: vp = ∂y ∂t = 5 × (200π) cos(200πt −1.57x) = 1000π cos(200πt −1.57x) The particle acceleration is: ap = ∂2y ∂t2 = −5 × (200π)2 sin(200πt −1.57x) = −(200π)2y Numerically: ap = −40000π2y ≈−3.95 × 105y m s−2 Thus, the acceleration of the particle is not zero. It is directly proportional to its displace- ment y and is directed toward the mean position (ap = −ω2y). It is zero only when the particle passes through the mean equilibrium position (y = 0) and reaches a maximum magnitude of ω2A ≈1.97 × 106 m/s2 at the crests and troughs (y = ±A). 3. Two progressive waves traveling in