JEE Physics Waves MCQs Set 08

Answer: (d) between 0 and 2A, depending on the phase difference

Explanation:
1. \( A_R=2A\cos\frac{\phi}{2} \) where \( \phi \) is the phase difference.
2. This ranges from 0 (φ = π) to 2A (φ = 0), so it lies between 0 and 2A.

Teacher's Note:
1. The resultant depends on the phase difference, not on amplitude alone.
2. Only the special cases φ = 0 and φ = π give exactly 2A and 0.

Answer: (d) the amplitude of the resultant depends on the phase angle while the frequency stays the same

Explanation:
1. Superposition of two waves of the same ω gives a wave of the same ω.
2. Amplitude \( 2A\cos(\phi/2) \) depends on φ.

Teacher's Note:
1. Superposition never changes the frequency of two waves having the same frequency.
2. Only the amplitude varies with phase difference.

Answer: (c) 2A and 8A

Explanation:
1. Incident and reflected waves travel in opposite directions with different amplitudes.
2. Node: \( 5A-3A=2A \); antinode: \( 5A+3A=8A \).

Teacher's Note:
1. A pure standing wave needs equal amplitudes; with unequal amplitudes the nodes do not become perfect zero.
2. Node amplitude is the difference, antinode amplitude the sum.

Answer: (b) \( y=0.5\cos(2\pi x+4\pi t) \)

Explanation:
1. Here \( k=\frac{2\pi}{\lambda}=2\pi\ \text{m}^{-1} \) and \( \omega=2\pi f=4\pi\ \text{rad/s} \).
2. A wave moving towards negative x has the same sign for the kx and ωt terms, so \( y=0.5\cos(2\pi x+4\pi t) \) fits.

Teacher's Note:
1. Same signs of kx and ωt mean the wave travels towards -x; opposite signs mean +x.
2. Always check that k and ω come out as 2π/λ and 2πf.

Answer: (a) \( 2\sqrt3\ \text{cm} \)

Explanation:
1. Amplitude at x = 5 cm: \( 4\sin\frac{5\pi}{15}=4\sin60^\circ=2\sqrt3\ \text{cm} \).

Teacher's Note:
1. The maximum displacement at a point is its amplitude, found by setting cos(wt) = 1.
2. Here sin 60 degrees equals root 3 over 2.

Answer: (a) \( 1.28\times10^{9}\ \text{N/m}^2 \)

Explanation:
1. \( v=2Lf=400\ \text{m/s} \).
2. \( v=\sqrt{T/\mu}=\sqrt{\text{stress}/\rho}\Rightarrow\text{stress}=\rho v^2=8000\times1.6\times10^5=1.28\times10^9\ \text{N/m}^2 \).

Teacher's Note:
1. Tension per unit area is the stress, and mu = rho A.
2. Wave speed on a wire is the square root of stress over density.

Answer: (d) \( \frac{32}{27} \)

Explanation:
1. \( f\propto\sqrt T \). In water \( T'=\frac T4=T\left(1-\frac1d\right)\Rightarrow d=\frac43 \) (d = relative density of the weights).
2. In the liquid \( T''=\frac T9=T\left(1-\frac sd\right)\Rightarrow\frac sd=\frac89\Rightarrow s=\frac89\times\frac43=\frac{32}{27} \).

Teacher's Note:
1. Buoyancy reduces the effective weight by the weight of the displaced liquid.
2. Frequency of a string is proportional to the square root of tension.

Answer: (c) h

Explanation:
1. Tension at height h is the weight of rope below: \( T=\mu gh \).
2. \( v^2=\frac{T}{\mu}=gh \), so \( v^2\propto h \) (speed increases as \( \sqrt h \)).

Teacher's Note:
1. The upper part of the rope supports more weight, so tension (and speed) grows upwards.
2. The speed-height graph is a parabola opening sideways, not a straight line.

Answer: (b) \( y=-0.8a\sin\frac{2\pi}{\lambda}(vt+x) \)

Explanation:
1. Reflection from a rigid surface inverts the wave (phase change π): negative sign.
2. The reflected wave travels in the -x direction: \( (vt+x) \), amplitude \( 0.8a \).

Teacher's Note:
1. A rigid boundary inverts the wave; a free end does not.
2. The direction reverses, so the sign between vt and x changes.

Answer: (b) \( 300\ \text{rad/s} \)

Explanation:
1. Both waves have the same coefficient of t, so ω = 300 rad/s.
2. Superposition of waves of equal ω gives a wave of the same ω.

Teacher's Note:
1. Read ω as the coefficient of t, and do not confuse it with frequency f = ω/2π.
2. The extra phase 0.2 only changes the amplitude and phase of the resultant.

Answer: (a) 3 cm

Explanation:
1. \( k=\frac{\pi}{3}\Rightarrow\lambda=\frac{2\pi}{k}=6\ \text{cm} \).
2. Adjacent nodes are \( \lambda/2=3\ \text{cm} \) apart.

Teacher's Note:
1. Distance between adjacent nodes (or antinodes) is lambda/2.
2. Compare with y = 2A sin(kx) cos(wt) to read off k.

Answer: (d) \( 2\pi/3 \)

Explanation:
1. \( 2a\cos\frac{\Delta\phi}{2}=a\Rightarrow\cos\frac{\Delta\phi}{2}=\frac12 \).
2. \( \frac{\Delta\phi}{2}=\frac{\pi}{3}\Rightarrow\Delta\phi=\frac{2\pi}{3} \).

Teacher's Note:
1. The resultant amplitude of two equal waves is 2a cos(delta/2).
2. Time period of the resultant is unchanged.

Answer: (c) 5

Explanation:
1. Phase difference \( \pi/2 \): \( A=\sqrt{4^2+3^2}=5 \).

Teacher's Note:
1. For a quarter-cycle phase difference the amplitudes add like perpendicular vectors.
2. The 3-4-5 triangle.

Answer: (c) \( 2A\cos\frac{\phi}{4} \)

Explanation:
1. Phase difference \( \delta=\phi/2 \).
2. \( A_R=2A\cos\frac{\delta}{2}=2A\cos\frac{\phi}{4} \).

Teacher's Note:
1. Resultant amplitude is 2A cos(delta/2) where delta is the true phase difference.
2. Here delta is already phi/2, so the angle inside the cosine is phi/4.

Answer: (b) Statement 1 is true; Statement 2 is false

Explanation:
1. Intensity \( \propto\omega^2A^2 \): \( I_1\propto\omega^2(2a)^2=4a^2\omega^2 \), \( I_2\propto(2\omega)^2a^2=4a^2\omega^2 \) (equal), so 1 is true.
2. Intensity also depends on frequency, so statement 2 is false.

Teacher's Note:
1. Intensity depends on both amplitude and frequency, each squared.
2. Doubling frequency and halving amplitude leaves intensity unchanged.

Answer: (a) 35.6 N

Explanation:
1. Four loops: \( \lambda=\frac{2L}{4}=0.45\ \text{m} \); \( v=f\lambda=27\ \text{m/s} \).
2. \( \mu=\frac{0.044}{0.9}=0.0489\ \text{kg/m} \), \( T=\mu v^2=0.0489\times729\approx35.6\ \text{N} \).

Teacher's Note:
1. The wavelength is twice the loop length.
2. Tension = mu v squared.

Answer: (d) \( 5.1\times10^{-4}\ \text{m} \)

Explanation:
1. \( a_{max}=\omega^2A\Rightarrow A=\frac{800}{(2\pi\times200)^2}=\frac{800}{1.58\times10^6}\approx5.1\times10^{-4}\ \text{m} \).

Teacher's Note:
1. The mid-point is an antinode, so its amplitude is the full amplitude of the standing wave.
2. Use omega = 2 pi f.

Answer: (d) -5.48 cm

Explanation:
1. Phase \( =-0.3\pi+0.25\pi=-0.05\pi \) rad.
2. \( y=0.350\sin(-0.05\pi)=-0.0548\ \text{m}=-5.48\ \text{cm} \).

Teacher's Note:
1. Substitute the values with the phase measured in radians.
2. The negative sign of the kx term with +10 pi t shows the wave travels towards +x.

Answer: (c) 5 cm

Explanation:
1. \( \mu=\frac{10^{-3}}{0.2}=0.005\ \text{kg/m},\ v=\sqrt{0.5/0.005}=10\ \text{m/s} \).
2. \( \lambda=\frac vf=0.1\ \text{m} \), node separation \( \lambda/2=5\ \text{cm} \).

Teacher's Note:
1. Node to node distance is half the wavelength.
2. Convert the mass to kilograms.

Answer: (b) 3

Explanation:
1. \( v=\sqrt{T/\mu}=\sqrt{360/(9\times10^{-3})}=200\ \text{m/s} \).
2. Successive resonances differ by \( \frac{v}{2L}=280-210=70\ \text{Hz} \).
3. Number of loops \( n=\frac{210}{70}=3 \).

Teacher's Note:
1. For a string fixed at both ends, the resonant frequencies are integer multiples of the fundamental.
2. The gap between consecutive resonances equals the fundamental frequency.

Answer: (a) 70 Hz

Explanation:
1. Consecutive harmonics differ by the fundamental frequency.
2. \( f_1=420-350=70\ \text{Hz} \) (350 Hz is the fifth harmonic).

Teacher's Note:
1. Check: 350 = 5 x 70 and 420 = 6 x 70.
2. This works for a string fixed at both ends, where all harmonics are present.

Answer: (a) 96%

Explanation:
1. \( \frac{A_i+A_r}{A_i-A_r}=1.5\Rightarrow\frac{A_r}{A_i}=0.2 \).
2. Reflected energy fraction \( =0.2^2=4\% \), so 96% is transmitted.

Teacher's Note:
1. Energy goes as amplitude squared.
2. Transmitted fraction = 1 minus reflected fraction.

Answer: (d) 50 Hz, 4.0 cm, 2.0 m/s

Explanation:
1. Quarter period = 5.0 ms, so \( T=20\ \text{ms},\ f=50\ \text{Hz} \).
2. Particles at the mean position are half a wavelength apart: \( \lambda=4.0\ \text{cm} \).
3. \( v=f\lambda=50\times0.04=2.0\ \text{m/s} \).

Teacher's Note:
1. Mean to extreme takes T/4.
2. Adjacent particles that are simultaneously at mean position are lambda/2 apart.

Answer: (c) 3

Explanation:
1. Add the phasors of unit amplitude: x-component \( 1+\frac12-\frac12-1=0 \), y-component \( 0+\frac{\sqrt3}{2}+\frac{\sqrt3}{2}+0=\sqrt3 \).
2. Resultant amplitude \( \sqrt3 \) times a single amplitude, so intensity \( =3I_0 \).

Teacher's Note:
1. Intensity is proportional to the square of the resultant amplitude.
2. Resolve the phasors into components rather than adding amplitudes directly.

Answer: (b) 60 cm/s

Explanation:
1. \( k=1\ \text{cm}^{-1},\ \omega=60\ \text{s}^{-1} \).
2. \( v=\frac\omega k=60\ \text{cm/s} \) (wavelength \( 2\pi \) cm, frequency \( 60/2\pi \) Hz).

Teacher's Note:
1. Wave speed is omega over k, and the units come from k in cm^-1.
2. Do not confuse omega with frequency.

Answer: (b) 25 kg

Explanation:
1. Fixed length and frequency: \( f=\frac{p}{2L}\sqrt{\frac{T}{\mu}} \) so \( p\sqrt T \) is constant.
2. \( 5\sqrt9=3\sqrt M\Rightarrow\sqrt M=5\Rightarrow M=25\ \text{kg} \).

Teacher's Note:
1. Fewer loops need a higher wave speed, so more tension.
2. The number of loops is inversely proportional to the square root of tension.

Answer: (a) 1 and 2 only

Explanation:
1. 1: amplitude \( 2\sin(\pi/6)=1\ \text{mm} \) (correct).
2. 2: \( v=200\pi\sin(\pi x)\cos(100\pi t)=200\pi(\tfrac12)\cos30^\circ=50\pi\sqrt3\approx157\sqrt3\ \text{mm/s} \) (correct).
3. 3: \( \lambda=2\ \text{cm} \), loop length 1 cm, so 10 cm contains 10 loops, not 5 (incorrect).

Teacher's Note:
1. Loop length = lambda/2 = pi/k.
2. Differentiate with respect to t to get particle velocity at a fixed x.