AIPMT 2009 and AIPMT 2008 Questions on Waves

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

E_x = 0;

E_y = 2.5 N/C cos[(2π×10⁶ rad/s)t – (π×10^–2 rad/m)x];

E_z = 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.

Two Kerala Engineering Entrance 2007 Questions on Waves

The following question appeared in Kerala Engineering Entrance (2007) Examination question paper. It has appeared in modified forms in other question papers as well:

A transverse wave is described by the equation y = y₀sin2π(ft – x/λ). The maximum particle velocity is equal to four times the wave velocity if

(a) λ = πy₀/4

(b) λ = πy₀/2

(c) λ = πy₀

(d) λ = 2πy₀

(e) λ = πy₀/3

The particle velocity is dy/dt = 2πfy₀ cos2π(ft – x/λ) and the maximum particle velocity is 2πfy₀ ( when the cosine term has its maximum value equal to one).

The wave velocity is fλ since f is the frequency and λ is the wave length.

When the maximum particle velocity is equal to four times the wave velocity, we have

2πfy₀ = 4fλ so that λ = πy₀/2

An open pipe (or open organ pipe) in Acoustics means a pipe open at both ends where as a closed pipe (or closed organ pipe) means a pipe closed at one end. The following question on organ pipes which appeared in Kerala Engineering Entrance 2007 Examination question paper high lights the difference between the fundamental frequencies as well as the overtones possible in open and closed organ pipes:

An open organ pipe is closed suddenly with the result that the second overtone of the closed pipe is found to be higher in frequency by 100 than the first overtone of the original pipe. Then the fundamental frequency of the open pipe is

(a) 200 s^–1

(b) 100 s^–1

(c) 300 s^–1

(d) 250 s^–1

(e) 150 s^–1

The closed end of a pipe is always a node where as the open end is always an antinode. In the fundamental mode, the frequency of vibration of the air column in the open pipe is such that the open ends are antinodes as usual, but they are consecutive antinodes. This means that the length of the open pipe is equal to half the wave length of the fundamental note produced by the pipe. But in a closed pipe in the fundamental mode, the closed end is a node and the open end is the next antinode so that the length of the closed pipe is equal to a quarter of the wave length of the fundamental note produced by the pipe. The fundamental frequency (n) of the closed pipe is therefore equal to half the fundamental frequency (N) of the open pipe of the same length.

Or, n = N/2

All harmonics (odd as well as even) are possible in open pipes where as odd harmonics only are possible in closed pipes. The first overtone of the closed pipe is the 3^rd harmonic and the second overtone is the 5^th harmonic.

Therefore, frequency of the second overtone of the closed pipe = 5n

The first overtone of the open pipe is its second harmonic and is equal to 2N

Therefore, we have

5n = 2N + 100

Since n = N/2, the above equation becomes

5 N/2 = 2N + 100 from which N = 200 s^–1

^{You will find similar questions from different branches of Physics at physicsplus.blogspot.com}