JEE Physics Waves MCQs Set 07

Answer: (a) \( \text{a wave moving in the }-x\text{ direction with speed }\sqrt{b/a} \)

Explanation:
1. \( ax^2+bt^2+2\sqrt{ab}xt=(\sqrt a\,x+\sqrt b\,t)^2 \), a function of \( x+\sqrt{b/a}\,t \).
2. So it is a pulse moving towards -x with speed \( \sqrt{b/a} \).

Teacher's Note:
1. Any function of (x + vt) moves toward -x with speed v.
2. Complete the square to see the combination.

Answer: (d) 20 units and 25 units

Explanation:
1. The two waves have equal amplitude and travel in opposite directions, forming a standing wave of maximum amplitude \( 10+10=20 \).
2. Wave number \( k=2\pi(0.02)\Rightarrow\lambda=50 \); loop length \( =\lambda/2=25 \).

Teacher's Note:
1. A loop is the portion between two consecutive nodes and has length lambda/2.
2. Maximum amplitude occurs at antinodes and equals the sum of the two amplitudes.

Answer: (c) \( y=A\sin\left(\frac{3\pi x}{L}\right)\cos\left(\frac{3\pi vt}{L}\right) \)

Explanation:
1. Two extra nodes mean 3 loops: \( L=\frac{3\lambda}{2}\Rightarrow\lambda=\frac{2L}{3} \), \( k=\frac{3\pi}{L} \), \( \omega=kv=\frac{3\pi v}{L} \).
2. A node at x = 0 requires \( \sin kx \).

Teacher's Note:
1. Same k in the space part and the time part gives omega = kv.
2. A fixed end at the origin means the spatial factor is sin(kx), not cos(kx).

Answer: (b) 5

Explanation:
1. Phasors at 0°, 90°, 180°: horizontal sum \( 10-7=3 \), vertical sum \( 4 \).
2. \( A=\sqrt{3^2+4^2}=5\ \mu\text{m} \).

Teacher's Note:
1. Add the three vectors by components, taking each successive phase step of 90 degrees.
2. The 3-4-5 triangle gives the answer directly.

Answer: (b) directly on the square of the wave amplitude and the square of the wave frequency

Explanation:
1. \( P=\frac12\mu v\omega^2A^2 \) with \( \omega=2\pi f \).
2. So \( P\propto A^2f^2 \).

Teacher's Note:
1. Power is proportional to omega squared and amplitude squared, for a given medium and speed.
2. Doubling the amplitude or the frequency quadruples the power.

Answer: (a) 5

Explanation:
1. \( -\cos\theta=\sin(\theta-90^\circ) \), so \( y_2=5\sin(\omega t-kx-240^\circ) \).
2. Phase difference \( 240^\circ\equiv120^\circ \) (cos 120° = cos 240°).
3. \( A_R=\sqrt{25+25+50\cos120^\circ}=5 \).

Teacher's Note:
1. Reduce both waves to sine form with the same argument before applying the phasor formula.
2. For equal amplitudes and a phase difference of 120 degrees the resultant has the same amplitude.

Answer: (d) \( -(2\ \text{mm})\sin(5t-40x) \)

Explanation:
1. Speeds: \( v_2=v_1/2 \). Reflection coefficient \( \frac{v_2-v_1}{v_2+v_1}=\frac{-1/2}{3/2}=-\frac13 \).
2. Reflected amplitude \( -2 \) mm, same wavelength and frequency, travelling in the +x direction: \( -2\sin(5t-40x) \) mm.

Teacher's Note:
1. Reflection at a heavier string gives a negative amplitude ratio (inversion).
2. The reflected wave travels back through the same string, so k and omega are unchanged.

Answer: (c) 11.0 m/s

Explanation:
1. \( v_{max}=A\omega=0.350\times10\pi=11.0\ \text{m/s} \).

Teacher's Note:
1. Maximum particle speed is amplitude times angular frequency.
2. It differs from the wave speed omega/k = 10 pi / 3 pi = 3.33 m/s.

Answer: (c) 300 Hz

Explanation:
1. Displacement \( \propto\sin\frac{n\pi x}{L} \); at \( x=\frac L4 \) this is \( \sin\frac{n\pi}{4} \) and it is maximum for \( n=2,6,10,\dots \).
2. Lowest mode is n = 2 (100 Hz, so \( f_1=50 \) Hz); next is n = 6: \( 6\times50=300\ \text{Hz} \).

Teacher's Note:
1. Odd harmonics n=1,3,5 have an antinode at the centre but not at L/4.
2. Set sin(n pi/4) equal to 1.

Answer: (b) 89%

Explanation:
1. With equal tension the impedances are in the ratio \( Z_2/Z_1=\sqrt4=2 \).
2. \( \frac{P_t}{P_i}=\frac{4Z_1Z_2}{(Z_1+Z_2)^2}=\frac{8}{9}\approx89\% \).

Teacher's Note:
1. Transmitted power fraction = 1 minus reflected power fraction; reflected fraction is (1/3) squared = 1/9.
2. Amplitude coefficient and power coefficient are different.

Answer: (a) \( \lambda=\frac{\pi Y_0}{2} \)

Explanation:
1. Maximum particle velocity \( =\omega Y_0=2\pi fY_0 \); wave velocity \( v=f\lambda \).
2. Set \( 2\pi fY_0=4f\lambda \), so \( \lambda=\frac{\pi Y_0}{2} \).

Teacher's Note:
1. Particle velocity (changes with time) and wave velocity (constant) are different things.
2. The frequency f cancels out, so the condition depends only on amplitude and wavelength.

Answer: (d) \( \frac{1}{300}\ \text{s} \)

Explanation:
1. At x = 4 m the phase is \( 8\pi-100\pi t+\frac{\pi}{3} \).
2. The displacement first becomes zero when \( \frac{\pi}{3}-100\pi t=0 \), giving \( t=\frac{1}{300}\ \text{s} \).

Teacher's Note:
1. Terms that are multiples of 2 pi (here 8 pi) can be dropped.
2. The displacement is zero when the phase is a multiple of pi; pick the smallest t > 0.

Answer: (d) 7.9 Hz

Explanation:
1. \( \mu=0.01\ \text{kg/m},\ v=\sqrt{250/0.01}=158\ \text{m/s} \).
2. \( f_1=\frac{v}{2L}=\frac{158}{20}=7.9\ \text{Hz} \) (next: 15.8 Hz, 23.7 Hz).

Teacher's Note:
1. Fundamental frequency is v over 2L for a string fixed at both ends.
2. Always convert grams to kilograms.

Answer: (c) \( \frac23A_i\cos(2k_1x-\omega_1t) \)

Explanation:
1. Same tension: \( v_2=v_1/2 \), so \( k_2=2k_1 \) (frequency is unchanged).
2. \( A_t=\frac{2v_2}{v_1+v_2}A_i=\frac{2(1/2)}{3/2}A_i=\frac23A_i \).

Teacher's Note:
1. Transmitted amplitude is 2 v2 /(v1 + v2) times the incident amplitude.
2. The transmitted wave keeps the same angular frequency but has a larger wave number in the heavier string.

Answer: (b) \( \pi/2 \)

Explanation:
1. \( y_2=b\sin(\omega t-kx+\pi/2) \).
2. So the phase difference is \( \pi/2 \).

Teacher's Note:
1. A cosine leads the sine of the same argument by a quarter of a cycle.
2. Phase difference does not depend on the amplitudes a and b.

Answer: (a) 80 N

Explanation:
1. \( \mu=0.2/4=0.05\ \text{kg/m} \); distance travelled \( =4\times8=32\ \text{m} \) in 0.8 s, so \( v=40\ \text{m/s} \).
2. \( T=\mu v^2=0.05\times1600=80\ \text{N} \).

Teacher's Note:
1. One trip back and forth is twice the wire length.
2. Tension equals mu times v squared.

Answer: (a) 6.25 N

Explanation:
1. \( v=\frac{\lambda}{T}=\frac{0.50}{0.04}=12.5\ \text{m/s} \).
2. \( T_{ension}=\mu v^2=0.04\times156.25=6.25\ \text{N} \).

Teacher's Note:
1. Read the period 0.04 s and wavelength 0.50 m directly from the equation.
2. Tension = mu v squared.

Answer: (d) \( \text{Its frequency is }60\ \text{Hz} \)

Explanation:
1. Same signs of 60t and 2x: the wave travels towards -x. \( v=\frac{60}{2}=30\ \text{m/s} \).
2. \( \lambda=\frac{2\pi}{2}=\pi\ \text{m} \) and amplitude \( 10^{-4} \) m are correct.
3. Frequency is \( \frac{60}{2\pi}=\frac{30}{\pi}\ \text{Hz} \), not 60 Hz.

Teacher's Note:
1. The number multiplying t is the angular frequency in rad/s, not the frequency in Hz.
2. Keep f = omega / 2 pi in mind.

Answer: (c) purely kinetic

Explanation:
1. Each pulse moves 4 cm in 2 s, so the two coincide exactly.
2. Upward and downward displacements cancel: the string is flat, so potential energy is zero. The energy is entirely kinetic.

Teacher's Note:
1. Energy is conserved, so the total cannot be zero even when displacement is zero everywhere.
2. At complete overlap of opposite pulses the string is straight but moving fastest.

Answer: (b) 60 Hz, 2.5 m

Explanation:
1. \( f=\frac{\omega}{2\pi}=\frac{120\pi}{2\pi}=60\ \text{Hz} \).
2. \( \lambda=\frac{2\pi}{k}=\frac{2\pi}{4\pi/5}=2.5\ \text{m} \).

Teacher's Note:
1. Compare with y = A sin(wt - kx) to read w and k.
2. Wavelength is 2 pi divided by k.

Answer: (b) 1, 3 and 4 only

Explanation:
1. Need \( \omega/k=100 \) and an antinode at x = 3 (\( \sin3k=\pm1 \)).
2. 1: \( k=\pi/6 \), \( \sin(\pi/2)=1 \), \( \omega/k=100 \) (valid). 2: \( k=\pi/3 \), \( \sin\pi=0 \) (node, invalid).
3. 3: \( k=5\pi/6 \), \( \sin(5\pi/2)=1 \), \( \omega/k=100 \) (valid). 4: \( k=5\pi/2 \), \( \sin(15\pi/2)=-1 \), \( \omega/k=100 \) (valid).

Teacher's Note:
1. The driven end behaves as an antinode while the fixed end is a node.
2. Both conditions, speed and end condition, must hold.

Answer: (a) \( \text{amplitude }a/2,\ \text{frequency }2n,\ \text{wavelength }\lambda/2 \)

Explanation:
1. \( \cos^2\theta=\frac{1+\cos2\theta}{2} \), so \( y=\frac a2+\frac a2\cos\left(4\pi nt-\frac{4\pi x}{\lambda}\right) \).
2. Amplitude a/2, \( \omega=4\pi n\Rightarrow f=2n \), \( k=4\pi/\lambda\Rightarrow\lambda'=\lambda/2 \).

Teacher's Note:
1. Use the double-angle identity to bring the square out.
2. The constant a/2 only shifts the mean level and is not part of the wave.

Answer: (d) The speed of the constituent waves is 4 m/s

Explanation:
1. \( k=10\pi\Rightarrow\lambda=0.2\ \text{m} \) (correct statement). \( \cos(10\pi\times0.3)=\cos3\pi=-1 \): antinode (correct). \( \cos(10\pi\times0.15)=\cos1.5\pi=0 \): node (correct).
2. Speed \( v=\frac\omega k=\frac{50\pi}{10\pi}=5\ \text{m/s} \), not 4 m/s, so that statement is wrong.

Teacher's Note:
1. Speed of the constituent progressive waves is omega divided by k.
2. Antinodes are where |cos kx| = 1, nodes where it is zero.

Answer: (c) 0.01 s

Explanation:
1. Tension \( T\approx Mg=10\ \text{N} \) (the string is very light).
2. \( v=\sqrt{T/\mu}=\sqrt{10/0.001}=100\ \text{m/s} \), so \( t=\frac{1}{100}=0.01\ \text{s} \).

Teacher's Note:
1. Speed on a string is sqrt(T/mu); do not forget that tension in a light string is almost constant.
2. Time = length divided by speed.

Answer: (c) \( 144^\circ \)

Explanation:
1. \( \omega=\frac{\pi}{4}\times4=\pi\ \text{rad/s} \).
2. \( \Delta\phi=\omega\Delta t=0.8\pi=144^\circ \).

Teacher's Note:
1. For one particle (fixed x) the phase change is omega times the time gap.
2. For two particles at one instant it would be k times the distance.

Answer: (b) \( 2\sqrt2\ \text{s} \)

Explanation:
1. \( v=\sqrt{gh} \), so \( t=\int_0^L\frac{dh}{\sqrt{gh}}=2\sqrt{\frac Lg}=2\sqrt2\ \text{s} \).

Teacher's Note:
1. The speed grows with height, so the time is not simply length over final speed.
2. The pulse accelerates with constant acceleration g/2.

Answer: (a) 0.16 m

Explanation:
1. Frequency stays the same, so \( \lambda\propto v\propto\sqrt{T} \) (the rope has uniform mass per length).
2. Bottom tension \( =5g=50 \) N (g = 10); top tension \( =20g=200 \) N.
3. \( \lambda_{top}=0.08\sqrt{200/50}=0.16\ \text{m} \).

Teacher's Note:
1. The frequency of a wave is fixed by the source and does not change when the wave moves along a medium.
2. Speed rises as tension rises up the rope, so the wavelength increases.