Today we will discuss the multiple choice questions on waves included in All India Pre-Medical / Pre-Dental Entrance Examination (AIPMT) 2009 and 2008 question papers. Here are the AIPMT 2009 questions:
(1) The electric field part of of an electromagnetic wave in a medium is represented by
Ex = 0;
Ey = 2.5 N/C cos[(2π×106 rad/s)t – (π×10–2 rad/m)x];
Ez = 0.
The wave is:
(1) moving along x-direction with frequency 106 Hz and wave length 100 m.
(2) moving along x-direction with frequency 106 Hz and wave length 200 m.
(3) moving along –x-direction with frequency 106 Hz and wave length 200 m.
(4) moving along y-direction with frequency 2π×106 Hz and wave length 200 m.
The electric field variation is in accordance with the equation y = A sin (ωt – kx) which represents a progressive wave proceeding along the positive x-direction. Instead of the usual displacement y we have the electric field E. In place of the angular frequency ω we have 2π×106 (remember ω = 2πn) which means that the linear frequency n is 106 Hz.
Since the propagation constant k = 2π/λ where λ is the wave length, we have
π×10–2 = 2π/λ from which λ = 200 ms–1.
So the correct option is (2).
[The units of electric field E (N/C), angular frequency ω (rad/s) and the propagation constant k (rad/m) given in the wave equation in the question should not distract you.
You should note that the negative sign in the wave equation y = A sin (ωt – kx) or y = A sin (kx – ωt) indicates that the wave is propagating along the positive x-direction. A wave propagating along the negative x-direction is represented by y = A sin (ωt + kx).
You can make use of any other form of the wave equation, for instance, y = A sin [2π(t/T – x/ λ)], also to solve the problem].
(2) A wave in a string has an amplitude of 2 cm. The wave travels in the + ve direction of x axis with a speed of 128 m/sec. and it is noted that 5 complete waves fit in 4 m length of the string. The equation describing the wave is:
(1) y = (0.02) m sin (15.7x − 2010t)
(2) y = (0.02) m sin (15.7x + 2010t)
(3) y = (0.02) m sin (7.85x − 1005t)
(4) y = (0.02) m sin (7.85x + 1005t)
The unit of amplitude is metre, shown by its symbol ‘m’ in the equation (which should not distract you).
The equation describing the wave propagating in the +ve x-direction is
y = A sin (kx – ωt)
The amplitude A as given in the question is 2 cm = 0.02 m.
The velocity of the wave, v = ω/k = 128 ms–1.
Wave length λ = 4/5 m.
But k = 2π/λ = 2π×5/4 = 7.85 m and so ω = kv = 7.85×128 = 1005
So the equation of the wave is
y = (0.02) m sin (7.85x − 1005t)
The correct option is (3).
Here is the AIPMT 2008 question:
The wave described by y = 0.25 sin(10πx – 2πt) where x and y are in metres and t in seconds is a wave traveling along the
(1) negative x-direction with amplitude 0.25 m and wavelength λ = 0.2 m.
(2) negative x-direction with frequency 1 Hz.
(3) positive x-direction with frequency π Hz and wavelength λ = 0.2 m.
(4) positive x-direction with frequency 1 Hz and wavelength λ = 0.2 m.
The given wave equation is in the form y = A sin [2π(x/λ – t/T)].
The negative sign in the equation shows that the wave is propagating along the positive x-direction.
By comparison we obtain 2πx/λ = 10πx from which λ = 0.2 m. Also, 2πt/T =2πt from which the frequency 1/T = 1 Hz. [Option (4)].
You will find a useful post on waves at AP Physics Resources.
