Download Torsional Pendulum - Higher Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! SOLUTION to T1R S2 2000 Question 1 (16 Marks) (a) The velocity is v = dx dt = 1.60(1.30) cos(1.30 t − 0.75)cm / s and the acceleration is a = dv dt = −1.60(1.30 )2 sin(1.30t − 0.75)cm2 / s = −(1.30)2 y At t=0 s Displacement: y = 1.60 sin( −0.75) = −1.60sin 43 o = −1.09cm Velocity: v = 2.08 cos( −43 o) = 1.52cm / s Acceleration a = −(1.30)2(1.09) = 1.84cm2 / s At t=0.60 s Displacement: y = 1.60 sin( 0.03) ≈1.60( 0.03) = 0.048 cm Velocity: v = 2.08cos(0.03) ≈ 2.08cm / s Acceleration a = −(1.30 )2 (0.048) = −0.081cm2 / s (b) A torsional pendulum has a period given by T = 2π I K In the question, a torque of 5Nm produces a deflection of 12° which gives a spring constant, K, given by K = external torque angular displacement = 5Nm 12o 2π 360 =23.9Nm/rad I = T 2π 2 K = 0.5s 2π 2 (23.9) = 0.151kgm2 Question 2 (18 Marks) Let source 1 emit waves in the positive x-direction such that y1 = y1m sin 2πν1(t − x1 v ) and source 2 emit in the negative x-direction: y2 = y2m sin 2πν2(t + x2 v ) We are told v=3m/s. x1, x2 are measured from source 1, source 2 respectively. Equating y1 at x1 = 0 to ys1 and y2 at x2 = 0 to ys2 we find y1m = 0.06m ν1 = ν2 = 0.5Hz and y2m = 0.02 m Superposition of the 2 waves at x1 = 12m and x2 = −8m gives y = y1 + y2 = 0.06sin π(t − 12 3 ) + 0.02 sin π(t − 8 3 ) = 0.06 sin πt + 0.02 sin( πt − 2π 3 ) = 0.06 sin πt + 0.02(sin πt cos 2π 3 − cos πt sin 2π 3 ) = 0.06 sin πt + 0.02[sin πt(− 1 2 ) − cos πt( 3 2 )] = 0.05sin πt − 0.0173cos πt (b) The sound (pressure) wave is given as p = 1.5sin{[ 2π λ ](x −330 t)} where m=1 giving λ1 = 600 nm (orange), is the only visible wavelength. The medals will appear blue/violet, since light reflected at the red/orange end of the spectrum will be attenuated and light from the blue/violet end of the spectrum will be most strongly reflected. Question 5 (14 Marks) (a) (i) The polariser reduces the incident intensity I0 to I = I0 2 . The analyser will transmit an intensity ′ I given by ′ I = I cos2 θ = I0 2 cos2 30 o = 0.375 I0 (ii) The polariser will transmit an intensity I = I0 cos 2θ = I0 cos 2 30o = 0.75I0 which is polarised at 30 degrees to the analyser’s axis, so that the final transmitted intensity is I f = I cos 2 θ = 0.75 I0 cos 2 30 o = 0.563 I0 (b) Quarter wave plate: For plate thickness l, the optical paths lengths of the ordinary and extraordinary rays are ln ⊥ and ln|| respectively. The two rays must emerge with a 90º phase difference so the optical paths must differ by (k + 1 4 )λ0 with k=0,1,2,3… The minimum thickness is given by λ0 4 = l(n⊥ − n| |) so that l = λ0 4(n⊥ − n| | ) = 589x10 −9 4(1.732-1.456) = 534x10−9 m = 534nm
