(c) (i) x0 = amplitude(ii) 1/2mv2(x02 – x2)
(iii)
PA04 Legacy question pack
3. (a) acceleration proportional to
displacement
directed to equilibrium position
(b) when x= A (the amplitude position)
(c) (i) x0 = amplitude
(ii) 1/2mv2(x02 – x2)
(iii)
4. (a) w = 37 x 10-3 m / 29
s = l D / w = 1.24mm
(b) l decreases from red to green therefore w decreases, fringe width decreases
(c) no diffraction so no dispersion, colours are not separated
6 (a) you need a 2nd piece of
Polaroid
rotate it through 90o
see if the light is blocked
(b) transverse
2 (a) (i) particle motion is perpendicular to wave velocity
(ii) f – same
A – same
phase – 3p/2
(iii) T = 2.5s, sketch position that is p out of phase
(b) (i) reflection has same
l as original
equal but opposite velocities
(ii) f – same
A – P is greater than Q
phase – p out of phase
(iii) sketch has3 nodes along the length of the string – since wavelength is halved.
3. (a) vmax = 2pfA = 0.79ms-1
(b) points B & C – where is comes to rest
(c)
4. (a) draw only slight curvature – with
same l as incident waves
(b) draw semi-circles – again with same
l as incident waves
(c) diffraction
5. (a) (i) nl - dsinq
d = 1/5 x 105
q
= sin-1(1 x 589 x 10-9 / (1/5 x 105))
(ii) tan17.1 = 0.2 / x
x = 0.2 / tan17.1 = 0.65m
(b) n = 2, find q as above then rearrange tanq = d / 0.65 to find d