Vibrating Strip - Higher Physics - Solved Past Paper, Exams of Physics

Download Vibrating Strip - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! SOLUTIONS 1231 T1 Q1. SHM Vibrating Strip (a)(i) For SHM, y = Asin(ω t + φ ) for amplitude A and angular frequency ω. Set φ = 0. (ii) The velocity is given by v = dy dx = ωA cosωt The maximum speed vm occurs when cos = 1, ∴ vm = ωA with ω = 2πυ , ν = 5Hz vm = 2π.5.(10.10−3)ms−1 = 0.314 ms− 1 The acceleration a in SHM is given by a = d2 y dt 2 = −ω 2Asin ωt The maximum value of the acceleration occurs when sin=1 with magnitude am = ω 2 ym = ω 2A ∴ am = ω 2A = (2πυ)2 A = (2π.5)2.10 −2 ms −2 = 9.87ms − 2 (b) Bead mass = 2 g, SHM frequency = 3 Hz. The acceleration (from part (a)) is am = ω 2A = (2πυ)2 A The downward force on the bead due to gravity, Fg , is Fg = mg = ( 2x10 −3 ).9.8 = 0.0196 N The bead will begin to lose contact when am ≥ Fg, or (2πν)2 A ≥ 0.0196 A ≥ 0.0196 (2π.3)2 A ≥ 5.5x10−5m = 0.055mm (i) Heavy damping (very viscous liquid) (ii) Light damping (moderate viscosity) Amplitude Amplitude x time time time xm e − at / m sin ωt xm e −bt/ m sin ωt (iii) Undamped motion (zero viscosity liquid) Amplitude x time xm sin ω t 3. Standing waves on string (i) The two waves given are y1 = 0.20 sin(2.0x − 4.0t) and y2 = 0.20sin(2.0x + 4.0t) and are of the form: y = ym sin( kx ± ωt) so that k = 2.0m −1 and ω = 4.0s−1 by inspection. Using the identity sin A + sin B = 2sin (A + B) 2 cos (A − B) 2 where A and B represent the two wave functions given, the standing wave is y1+ 2 given by y1+2 = 2ym sin kxcosωt = 0.40sin(2.0x)cos(4.0t) (ii) At position x = 0.45m y = 0.40sin( 0.90)cos(4.0t) = 0.31cos(4.0t) ∴ maximum amplitude with value y=0.31m occurs when cos(4.0t)=1 (iii) For the standing wave pattern y1+2 = 0.40 sin( 2.0x) cos(4.0t) we will have nodes at both ends of the string. For such a string fixed at both ends, nodes are also located at positions x = n λ 2 , so that λ 2 = 1 2 2π k = π 2.0 m = 1.57m A standing wave will result when the other end of the string is fixed at x position x = n(1.57m) = 1.57m, 3.14m,...... (n=1,2,3….) ∴ ∆θ = θ2 − θ1 = λ d = 600x10 −9 0.5x10−3 = 1.2x10−3rad The linear distance between fringes on the screen will be δ = L∆θ = (1m)x(1.2x10−3rad) = 1.2mm (b) The intensity pattern on the screen is the product of the interference effect modulated by the diffraction ‘envelope’. Diffraction minima occur according to the condition asin θ = nλ (n integer) where a is the slit width. For n=1, θ = λ d = 600 x10 −9 0.1x10 −3 = 6x10 −3rad The 5th interference fringe in this pattern has zero intensity – it is ‘modulated’ to zero by the diffraction envelope. The pattern looks like: (c) The intensity envelope arises because diffraction at each slit modulates the ‘strength’ (intensity) of the interference fringes. Using a phasor diagram: The slit pattern has N sub-rays each differing in phase φ by φ = 2πa λ sin θ for slit width a, angular position on screen θ . Referring to the diagram, φ = Em R and Eθ = 2Rsin φ 2 = Em φ 2 sin φ 2       = Em sinφ/ 2 φ /2           2 and the intensity is Iθ = Im sin φ / 2 φ / 2           2 The intensity of the 3rd fringe relative to the central maximum is Iθ = Im cos 2 β sin φ/ 2 φ/ 2           where β = πd λ sin θ = 3π (d sin θ = 3λ ⇒ sin θ = 3.6x10 −3 ) φ / 2 = πa λ sin θ = πa λ (3.6x10−3) = 0.6π ∴ I3 = Im cos 2 (3π ) sin 0.6π 0.6π       2 = 0.255 Im (d) If there are four (rather than two) slits of equal width, the width of the interference fringes is reduced according to ∆θ = λ Nd where ∆θ is the angular width of interference fringes, λ is the wavelength of illuminating light and N is the number of slits. In this case, doubling the number of (equivalent) slits halves the fringe widths to new value 0.3x10-3 rad.