JEE Physics Waves MCQs Set 14

Answer: (b) 2

Explanation:
1. Waves travelling in opposite directions produce a stationary wave, in which the space part and the time part are in separate factors.
2. Equations (1) and (3) are functions of \( (kx\mp\omega t) \), so they are travelling waves.
3. Equation (2) has the form \( 2A\sin kx\cos\omega t \), a stationary wave, formed by two waves travelling in opposite directions.

Teacher's Note:
1. A stationary wave has the form f(x) g(t); a travelling wave depends on x ± vt.
2. Two equal waves in opposite directions give a stationary wave.

Answer: (d) 7, 6

Explanation:
1. The n-th harmonic of a string fixed at both ends has n loops, so n antinodes.
2. The number of nodes (including the two fixed ends) is \( n+1 \).
3. For n = 6: 7 nodes and 6 antinodes.

Teacher's Note:
1. Each loop contains one antinode and is bounded by two nodes.
2. Count the fixed ends as nodes.

Answer: (d) 9

Explanation:
1. The number of nodes equals the number of antinodes, so one end of the pipe is closed (a node) and the other is open (an antinode).
2. In a closed pipe the harmonics are odd: \( n=2N-1 \) for N nodes (or antinodes).
3. With N = 5, \( n=2\times5-1=9 \).

Teacher's Note:
1. A pipe closed at one end has a node at the closed end and an antinode at the open end.
2. Fundamental: 1 node and 1 antinode; third harmonic: 2 and 2; and so on.

Answer: (a) open pipe

Explanation:
1. A pipe open at both ends supports all harmonics (1, 2, 3, ...).
2. A pipe closed at one end supports only odd harmonics (1, 3, 5, ...).
3. The tenth harmonic is even, so the pipe must be open.

Teacher's Note:
1. Even harmonics are present only in open pipes.
2. A closed pipe is characterised by odd harmonics only.

Answer: (c) 150, 450 Hz

Explanation:
1. The frequencies 300, 600, 750 and 900 are all multiples of 150 Hz, and 750 = 5 × 150 shows that both even and odd multiples occur, so the tube is open at both ends with fundamental 150 Hz.
2. Harmonics below 1000 Hz: 150, 300, 450, 600, 750, 900 (six in all).
3. The missing ones are 150 Hz and 450 Hz.

Teacher's Note:
1. The highest common factor of the given frequencies gives the fundamental.
2. An open pipe has every integer multiple of the fundamental.

Answer: (b) 292 cps

Explanation:
1. The unknown fork has frequency \( 288\pm4=284\ \text{or}\ 292\ \text{cps} \).
2. Loading with wax lowers its frequency. If it were 284 Hz it would drop further and the beats would increase above 4. Since the beats decrease to 2, the unknown fork must have been higher than 288 Hz, and wax brought it to 290 Hz.
3. So its frequency was 292 cps.

Teacher's Note:
1. Wax increases the mass of the prong and so lowers the frequency.
2. Beats decreasing after lowering the frequency means the fork was above the reference fork.

Answer: (b) \( y=-a\cos(kx+\omega t) \)

Explanation:
1. A stationary wave needs a wave of the same amplitude travelling in the opposite direction, so the other wave has \( (kx+\omega t) \).
2. Take \( y_2=-a\cos(kx+\omega t) \). Then \( y=a\cos(kx-\omega t)-a\cos(kx+\omega t)=2a\sin kx\sin\omega t \).
3. At x = 0, y = 0 for all times, so x = 0 is a node, as required.

Teacher's Note:
1. A wave reflected from a fixed end has a phase change of π, which gives the minus sign.
2. The sum 2a sin kx sin ωt vanishes at x = 0.

Answer: (d) when two identical waves with a phase difference of π are moving in the same direction

Explanation:
1. Standing waves arise from the superposition of two waves of the same frequency travelling in opposite directions (for example an incident wave and its reflection).
2. Options (a), (b) and (c) all involve reflection, so they produce standing waves.
3. Two identical waves moving in the same direction, differing in phase by π, simply cancel each other; no stationary pattern is formed.

Teacher's Note:
1. Opposite directions of travel are essential for a standing wave.
2. Waves moving in the same direction add up to another travelling wave.

Answer: (a) 1.21 Å

Explanation:
1. Nodes and antinodes alternate. With 3 nodes and 2 antinodes the pattern runs from node to node through two loops.
2. Each loop has length \( \frac\lambda2 \), so \( 2\times\frac\lambda2=1.21\ \text{Å} \).
3. \( \lambda=1.21\ \text{Å} \).

Teacher's Note:
1. Two adjacent nodes are λ/2 apart.
2. Number of loops = number of antinodes.

Answer: (d) 10 m

Explanation:
1. The two waves form a standing wave \( y=2A\sin kx\cos\omega t \).
2. The string is flat whenever \( \cos\omega t=0 \), which happens twice per period, so the minimum interval between flat instants is \( \frac T2=0.5\ \text{s}\Rightarrow T=1\ \text{s} \).
3. \( \lambda=vT=10\times1=10\ \text{m} \).

Teacher's Note:
1. Flat means all displacements are zero at the same moment.
2. In one time period there are two such instants.

Answer: (c) 5 Hz

Explanation:
1. Five segments means five loops: \( 5\times\frac\lambda2=10\Rightarrow\lambda=4\ \text{m} \).
2. \( f=\frac v\lambda=\frac{20}{4}=5\ \text{Hz} \).

Teacher's Note:
1. Each segment (loop) has length λ/2.
2. This is the fifth harmonic of a string with fundamental 1 Hz.

Answer: (c) 40 cm

Explanation:
1. The fundamental frequency of a fixed string is \( f=\frac{1}{2L}\sqrt{T/\mu} \), so \( f\propto\frac1L \) at fixed tension.
2. \( L_2=L_1\frac{f_1}{f_2}=50\times\frac{800}{1000}=40\ \text{cm} \).

Teacher's Note:
1. Shortening a string raises its pitch.
2. Tension and mass per unit length are unchanged.

Answer: (a) 6

Explanation:
1. Train coming in: \( f_1=240\times\frac{320}{320-4}=243.04\ \text{Hz} \).
2. Train leaving: \( f_2=240\times\frac{320}{320+4}=237.04\ \text{Hz} \).
3. Beats per second \( =f_1-f_2=6 \).

Teacher's Note:
1. The approaching train raises the pitch and the receding one lowers it.
2. The beat frequency is the difference between the two heard frequencies.

Answer: (c) \( n=5,\ f_2=\frac54f_1 \)

Explanation:
1. Open pipe, 2nd harmonic: \( f_1=\frac{2v}{2L}=\frac vL \).
2. Closed pipe frequencies: \( f=(2m-1)\frac{v}{4L}=(2m-1)\frac{f_1}{4} \): \( \frac{f_1}{4},\frac{3f_1}{4},\frac{5f_1}{4},\dots \).
3. The first one above \( f_1 \) is \( \frac54f_1 \), the 5th harmonic of the closed pipe (n = 5).

Teacher's Note:
1. Closing one end halves the fundamental and keeps only odd harmonics.
2. Compare each closed-pipe resonance with f1 to find the first one higher than f1.

Answer: (c) 6

Explanation:
1. A closed pipe produces odd harmonics: \( 1500,\ 4500,\ 7500,\ \dots,\ 1500(2n-1) \).
2. Audibility limit: \( 1500(2n-1)\le20000\Rightarrow2n-1\le13.3 \), so the odd harmonics are 1, 3, 5, 7, 9, 11, 13.
3. That is 7 harmonics, of which the first is the fundamental, leaving 6 overtones.

Teacher's Note:
1. Overtones are the harmonics above the fundamental.
2. The highest audible frequency is 20 kHz.

Answer: (d) \( \sqrt2A:\omega \)

Explanation:
1. Resultant amplitude: \( A_R=\sqrt{A^2+A^2+2A^2\cos\frac\pi2}=\sqrt2A \).
2. Superposition of waves of the same frequency does not change the frequency, so it remains ω.

Teacher's Note:
1. Use the phasor addition formula for the amplitude.
2. Waves of the same frequency give a resultant of the same frequency.

Answer: (a) 5

Explanation:
1. The phase difference is \( \frac\pi2 \), so the phasors are perpendicular.
2. \( A=\sqrt{4^2+3^2+2(4)(3)\cos\frac\pi2}=\sqrt{25}=5 \).

Teacher's Note:
1. With 90° between them, the amplitudes add like the sides of a right triangle.
2. This is the familiar 3-4-5 triangle.

Answer: (a) A and B

Explanation:
1. A stationary wave requires two waves of the same amplitude and frequency travelling along the same line in opposite directions.
2. A travels in the +x direction and B in the -x direction, so together they give a stationary wave.
3. C travels along the y-direction, so it does not form a stationary pattern with A or B.

Teacher's Note:
1. Check the direction of travel from the sign between kx and ωt and from which coordinate appears.
2. Waves along different axes do not produce standing waves.

Answer: (a) \( 10^4\ \text{m/s} \)

Explanation:
1. Compare with \( y=2a\sin\omega t\cos kx \): \( \omega=100\ \text{rad/s} \) and \( k=0.01\ \text{rad/m} \).
2. The velocity of the component waves is \( v=\frac\omega k=\frac{100}{0.01}=10^4\ \text{m/s} \).

Teacher's Note:
1. The speed of the component travelling waves is ω/k, even though the standing wave itself does not travel.
2. Units: the x coordinate is in metres, so k is in rad/m.

Answer: (b) \( \frac{P_A+2P_0}{P_A-2P_0} \)

Explanation:
1. At the closed end there is a displacement node, which is a pressure antinode.
2. The incident and the reflected waves arrive in phase there (in terms of pressure), each of amplitude \( P_0 \), so the pressure amplitude of the standing wave at the closed end is \( 2P_0 \).
3. The pressure oscillates between \( P_A+2P_0 \) and \( P_A-2P_0 \), so the ratio is \( \frac{P_A+2P_0}{P_A-2P_0} \).

Teacher's Note:
1. A displacement node is a pressure antinode, where the incident and reflected pressure waves add.
2. Pressure oscillates about atmospheric pressure by plus or minus the amplitude.

Answer: (a) 144 cm

Explanation:
1. For the n-th mode the node spacing is \( \frac\lambda2=\frac Ln \).
2. Two consecutive modes: \( \frac Ln=18 \) and \( \frac L{n+1}=16 \).
3. So \( 18n=16(n+1)\Rightarrow n=8 \) and \( L=18\times8=144\ \text{cm} \).

Teacher's Note:
1. The higher mode has the smaller node spacing.
2. The length must be a whole number of half-wavelengths for each mode.

Answer: (d) none of these

Explanation:
1. One octave doubles the frequency, so two octaves multiply it by 4.
2. \( 4\times128=512\ \text{Hz} \), which is none of the listed values.

Teacher's Note:
1. An octave corresponds to a frequency ratio of 2 : 1.
2. 256 Hz is only one octave higher.

Answer: (d) Sound wave in a gas

Explanation:
1. X-rays, gamma rays and visible light are electromagnetic waves; the electric and magnetic fields oscillate perpendicular to the direction of propagation, so they are transverse.
2. A gas cannot sustain shear, so sound in a gas is a longitudinal compression wave.

Teacher's Note:
1. Fluids support only longitudinal waves because they have no rigidity.
2. All electromagnetic waves are transverse.

Answer: (d) infrasonic waves

Explanation:
1. For the same speed, a larger wavelength means a smaller frequency, \( f=v/\lambda \).
2. Audible sound is in the range 20 Hz to 20 kHz, so waves of longer wavelength have a frequency below 20 Hz: infrasonic waves.

Teacher's Note:
1. Wavelength and frequency are inversely related at a fixed speed.
2. Ultrasonic waves have frequencies above 20 kHz, hence shorter wavelengths.

Answer: (b) Ultrasonic waves

Explanation:
1. SONAR (sound navigation and ranging) sends pulses of high-frequency sound into water and times the echo.
2. Ultrasonic waves (above 20 kHz) are used because they have short wavelengths, can be sent as narrow beams and are not heard by people.

Teacher's Note:
1. Radio and light waves are absorbed strongly by water, so they are not used for underwater ranging.
2. Distance = (speed of sound in water × echo time) / 2.