JEE Physics Waves MCQs Set 17

Answer: (a) wave speed v

Explanation:
1. AB is one wavelength: \( AB=\lambda \).
2. CD is one time period: \( CD=T \).
3. \( \frac{AB}{CD}=\frac{\lambda}{T}=v \), the wave speed.

Teacher's Note:
1. A y-x graph gives the wavelength and a y-t graph gives the time period.
2. Wavelength divided by time period is always the wave speed.

Answer: (b) 80 N

Explanation:
1. Comparing with \( y=A\sin(\omega t-kx) \): \( \omega=420\ \text{rad/s},\ k=21\ \text{rad/m} \).
2. \( v=\frac{\omega}{k}=\frac{420}{21}=20\ \text{m/s} \)
3. \( T=\mu v^2=0.200\times400=80\ \text{N} \)

Teacher's Note:
1. The wave speed on a string is \( v=\sqrt{T/\mu} \).
2. Read ω and k directly from the wave equation.

Answer: (c) \( 300\left(\frac{2\rho-1}{2\rho}\right)^{1/2} \)

Explanation:
1. Let V be the volume of the object; in air the tension is \( T_1=\rho Vg \) (water density taken as 1).
2. Half submerged: buoyancy is \( \tfrac12Vg \), so \( T_2=\left(\rho-\tfrac12\right)Vg \).
3. \( f\propto\sqrt{T}\Rightarrow f'=300\sqrt{\frac{\rho-1/2}{\rho}}=300\left(\frac{2\rho-1}{2\rho}\right)^{1/2} \)

Teacher's Note:
1. Buoyancy reduces the tension in the wire.
2. Fundamental frequency of a wire of fixed length is proportional to the square root of the tension.

Answer: (d) \( \dfrac{\omega L}{\pi} \)

Explanation:
1. Speed of pulse relative to rope: \( \omega R=\frac{\omega L}{2\pi} \); speed of the rope itself: \( \omega R \).
2. Same direction: \( v_g=\omega R+\omega R=2\omega R=\frac{\omega L}{\pi} \)

Teacher's Note:
1. Add the pulse speed and the rope speed when both are in the same direction.
2. Subtract them when they are opposite.

Answer: (d) \( y=\log(x^2-t^2)-\log(x-t) \)

Explanation:
1. A travelling wave must be a function of \( x\pm vt \) only.
2. \( \log(x^2-t^2)-\log(x-t)=\log\frac{(x-t)(x+t)}{x-t}=\log(x+t) \), a function of \( x+t \).
3. The others are not functions of a single combination of x and t (and do not satisfy the wave equation).

Teacher's Note:
1. Factorise the argument of the logarithm before deciding.
2. A function of x + vt travels in the negative x direction.

Answer: (d) 144 m/s

Explanation:
1. The two frequencies are consecutive harmonics, so the fundamental is \( f_1=384-288=96\ \text{Hz} \).
2. \( v=2Lf_1=2\times0.75\times96=144\ \text{m/s} \)

Teacher's Note:
1. For a string fixed at both ends, resonant frequencies are integral multiples of the fundamental.
2. The difference between consecutive harmonics equals the fundamental.

Answer: (b) The pulses pass through each other without any change in their shape

Explanation:
1. Waves obey the principle of superposition: displacements add while the pulses overlap.
2. After overlapping, each pulse continues with its original shape and speed.

Teacher's Note:
1. Superposition is temporary; it does not change the pulses afterwards.
2. This is the basic difference between waves and colliding particles.

Answer: (a) Height 2 cm from x = -1 to 0; rising linearly from 2 cm at x = 0 to 4 cm at x = 1; then dropping to 2 cm at x = 1 and falling linearly to 0 at x = 2 (cm)

Explanation:
1. In 2 s each pulse moves 1 cm. The rectangle now spans x = -1 to 1 cm and the triangle spans x = 0 to 2 cm with its peak at x = 1 cm.
2. For x from -1 to 0 only the rectangle is present: displacement 2 cm.
3. For x from 0 to 1 the triangle rises from 0 to 2 cm and adds to the 2 cm of the rectangle: displacement rises from 2 cm to 4 cm.
4. For x from 1 to 2 only the triangle remains, falling from 2 cm to 0. The resultant is therefore the first shape.

Teacher's Note:
1. Draw each pulse after it has moved by speed × time, then add the displacements point by point.
2. The rectangle is 2 cm tall everywhere in its width, so it adds a constant 2 cm.

Answer: (b) the speed of the component waves

Explanation:
1. The standing wave is the superposition of two waves \( A\sin(\omega t\mp kx) \) travelling in opposite directions.
2. Each has speed \( v=\omega/k \), which depends on tension and linear mass density.

Teacher's Note:
1. A standing wave itself has no translational speed.
2. The wave speed on a string is √(T/μ).

Answer: (b) 35 Hz

Explanation:
1. Strain \( =\frac{0.05\ \text{cm}}{100\ \text{cm}}=5\times10^{-4} \), so stress (tension per unit area) \( =Y\times\text{strain}=4.5\times10^{7}\ \text{N/m}^2 \).
2. \( v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{\text{stress}}{\rho}}=\sqrt{\frac{4.5\times10^7}{9000}}=70.7\ \text{m/s} \)
3. \( f=\frac{v}{2L}=\frac{70.7}{2}\approx35\ \text{Hz} \)

Teacher's Note:
1. T/μ equals stress divided by density because μ = ρA.
2. Convert 9 g/cm³ to 9000 kg/m³.

Answer: (c) both the speed of propagation and the wavelength

Explanation:
1. Frequency is fixed by the source and does not change at the junction.
2. The linear densities differ, so \( v=\sqrt{T/\mu} \) changes.
3. With the same frequency and a new speed, \( \lambda=v/f \) also changes.

Teacher's Note:
1. Tension is the same on both sides; linear density differs.
2. Frequency never changes when a wave crosses a boundary.

Answer: (c) 2

Explanation:
1. Frequency does not change, so \( \lambda\propto v\propto\sqrt{T} \).
2. At the bottom \( T_b=mg \); at the top \( T_t=(m+M)g=2mg \).
3. \( a=\frac{\lambda_t}{\lambda_b}=\sqrt{2}\Rightarrow a^2=2 \)

Teacher's Note:
1. Tension at the top supports both the block and the whole rope.
2. Wavelength changes in proportion to speed because frequency is constant.

Answer: (b) 8

Explanation:
1. \( M=\int_0^L kx\,dx=\frac{kL^2}{2}\Rightarrow k=\frac{2M}{L^2} \).
2. \( t=\int_0^L\frac{dx}{v}=\int_0^L\sqrt{\frac{kx}{F}}dx=\frac23\sqrt{\frac{k}{F}}L^{3/2}=\frac23\sqrt{\frac{2ML}{F}}=\sqrt{\frac{8ML}{9F}} \)
3. Comparing, p = 8.

Teacher's Note:
1. Integrate the time dt = dx/v along the rope.
2. Use the total mass to eliminate k.

Answer: (d) 15

Explanation:
1. The hand crosses the mean position twice per cycle, so \( f=3\ \text{Hz},\ T=\tfrac13\ \text{s} \).
2. The disturbance takes \( \frac{3}{3}=1\ \text{s} \) to reach the point.
3. Starting downward from the mean, the hand reaches the upper extreme after \( \tfrac34T=\tfrac14\ \text{s} \).
4. \( t=1+\frac14=\frac54=\frac{15}{12}\ \text{s} \), so n = 15.

Teacher's Note:
1. Passing the mean position twice per cycle gives the frequency.
2. The upper extreme is reached three-quarters of a period after starting downward.

Answer: (c) 5

Explanation:
1. Fundamental of left wire = first overtone (second harmonic) of right wire: \( \frac{1}{2\ell}\sqrt{\frac{T_1}{\mu}}=\frac{2}{2\ell}\sqrt{\frac{T_2}{\mu}}\Rightarrow T_1=4T_2 \).
2. Total weight \( (1.2+4.8)\times10=60\ \text{N} \), so \( T_1=48\ \text{N},\ T_2=12\ \text{N} \).
3. Torque about the left end: \( 12\times40=12\times20+48x\Rightarrow x=5\ \text{cm} \)

Teacher's Note:
1. Frequency of the second harmonic is twice that of the fundamental.
2. Take moments about the left end to find x.

Answer: (b) 3

Explanation:
1. Third overtone is the fourth harmonic: \( A(x)=a\left|\sin\frac{4\pi x}{\ell}\right| \).
2. At \( x=\ell/3 \): \( A=a\sin\frac{4\pi}{3}\to\frac{\sqrt3}{2}a \).
3. So \( \sqrt p=\sqrt3 \) and p = 3.

Teacher's Note:
1. Third overtone means the fourth harmonic.
2. Use the magnitude of the sine.

Answer: (c) 4

Explanation:
1. Antinode amplitude (waves in phase): \( 5A+3A=8A \).
2. Node (minimum) amplitude (waves out of phase): \( 5A-3A=2A \).
3. Ratio \( \frac{8A}{2A}=4 \).

Teacher's Note:
1. Maximum amplitude is the sum of the two amplitudes; minimum is their difference.
2. The ratio of the two is the standing wave ratio.

Answer: (b) 80

Explanation:
1. Light string at the far end allows that end to move freely, so the rope behaves as a string fixed at one end only.
2. Second overtone is the fifth harmonic: \( L=\frac{5\lambda}{4}\Rightarrow\lambda=\frac{4\times100}{5}=80\ \text{cm} \).
3. (Speed \( v=\sqrt{T/\mu}=\sqrt{400/0.04}=100\ \text{m/s} \) is not needed for the wavelength.)

Teacher's Note:
1. For one end fixed and one free only odd harmonics are present.
2. The second overtone is the 5th harmonic.

Answer: (d) 36

Explanation:
1. \( \mu=\frac{0.044}{0.90}=0.0489\ \text{kg/m} \).
2. Four loops: \( L=4\cdot\frac{\lambda}{2}\Rightarrow\lambda=0.45\ \text{m} \), so \( v=f\lambda=60\times0.45=27\ \text{m/s} \).
3. \( T=\mu v^2=0.0489\times729=35.6\approx36\ \text{N} \)

Teacher's Note:
1. Four loops means L = 2λ.
2. T = μv².

Answer: (b) 5

Explanation:
1. At fixed frequency and length, \( n\sqrt{\mu/T}=\text{constant} \), so \( 10\sqrt{T_a}=11\sqrt{T_w} \).
2. \( \frac{T_w}{T_a}=\frac{d-1}{d}=\left(\frac{10}{11}\right)^2=0.826\Rightarrow d=\frac{121}{21}=5.76 \).
3. The greatest integer not exceeding 5.76 is 5.

Teacher's Note:
1. The harmonic number is inversely proportional to the square root of the tension.
2. Buoyancy lowers the tension to (d − 1)/d of its original value.

Answer: (a) all four statements (i), (ii), (iii) and (iv)

Explanation:
1. \( \omega=60,\ k=2\Rightarrow v=\frac{\omega}{k}=30\ \text{m/s} \); the sign of \( kx \) and \( \omega t \) is the same, so the wave travels along \( -x \).
2. \( \lambda=\frac{2\pi}{k}=\pi\ \text{m} \).
3. \( f=\frac{\omega}{2\pi}=\frac{30}{\pi}\ \text{Hz} \).
4. Amplitude is the coefficient \( 10^{-4} \) m. All four are correct.

Teacher's Note:
1. Same sign between ωt and kx means travel in the negative x direction.
2. Check each statement from ω = 60 and k = 2.

Answer: (b) (ii), (iii) and (iv) only

Explanation:
1. Here \( \omega=\alpha \) and \( k=\beta \).
2. \( f=\frac{\omega}{2\pi}=\frac{\alpha}{2\pi} \) (so (i) is wrong), \( \lambda=\frac{2\pi}{k}=\frac{2\pi}{\beta} \) and \( v=\frac{\omega}{k}=\frac{\alpha}{\beta} \).

Teacher's Note:
1. α is the angular frequency, not the frequency.
2. β is the wave number.

Answer: (b) (i) and (iii) only

Explanation:
1. (i) is a product that satisfies the wave equation with \( v=\omega/k \): a standing wave.
2. (iii) is a function of \( kx+\omega t \) alone, so it is a travelling wave.
3. (ii) gives \( y_{tt}=-2\omega^2 \) and \( y_{xx}=2k^2 \), which cannot satisfy \( y_{tt}=v^2y_{xx} \). Similarly (iv) does not satisfy it.

Teacher's Note:
1. A wave function should satisfy ∂²y/∂t² = v² ∂²y/∂x².
2. A function of (kx + ωt) only is always a travelling wave.

Answer: (b) (i) and (iii) only

Explanation:
1. \( f\propto v\propto\sqrt T\Rightarrow\frac{f'}{f}=\sqrt{1.21}=1.1 \): speed and frequency each increase by 10 per cent, so (iii) is true and (ii) is false.
2. \( 0.1f=15\Rightarrow f=150\ \text{Hz} \), so (i) is true.
3. The length is unchanged, so the fundamental wavelength 2L does not change: (iv) is false.

Teacher's Note:
1. Fundamental frequency is proportional to the square root of the tension.
2. Wavelength of the fundamental is fixed by the string length.

Answer: (c) (iii) and (iv) only

Explanation:
1. Maxima: \( 3A+A=4A \); minima: \( 3A-A=2A \) (not zero), so (i) is false.
2. The maximum-to-adjacent-minimum distance is \( \frac{\lambda}{4}=\frac{v}{4f} \), so (ii) is false.
3. The resultant is a standing wave plus a travelling wave, both of which average to zero over a period, so (iii) is true; the amplitude pattern is fixed in space, so (iv) is true.

Teacher's Note:
1. Unequal amplitudes cannot cancel completely.
2. Adjacent antinode and node are a quarter wavelength apart.

Answer: (a) all four statements (i), (ii), (iii) and (iv)

Explanation:
1. Nodes where \( \cos10\pi x=0 \): \( 10\pi x=\left(n+\tfrac12\right)\pi\Rightarrow x=0.05+0.1n \), which includes 0.15 m.
2. Antinodes where \( |\cos10\pi x|=1 \): \( x=0.1n \), which includes 0.3 m.
3. \( \lambda=\frac{2\pi}{10\pi}=0.2\ \text{m} \) and \( v=\frac{\omega}{k}=\frac{50\pi}{10\pi}=5\ \text{m/s} \).

Teacher's Note:
1. Nodes are at odd multiples of λ/4.
2. The speed of the component waves is ω/k.