JEE Physics Waves MCQs Set 16

Answer: (d) \( \sqrt2\,y_m \)

Explanation:
1. \( A=2y_m\cos\frac{\phi}{2}=2y_m\cos\frac{\pi}{4}=\sqrt2\,y_m \)

Teacher's Note:
1. Two equal waves with phase difference φ superpose to amplitude 2y_m cos(φ/2).
2. Check with phasors: two perpendicular vectors of length y_m give a resultant of √2 y_m.

Answer: (b) 3.77 m/s

Explanation:
1. \( \omega=2\pi f=10\pi\ \text{rad/s} \).
2. \( v_{max}=A\omega=0.12\times10\pi=3.77\ \text{m/s} \)

Teacher's Note:
1. Transverse particle speed is different from the wave speed.
2. Maximum particle speed is amplitude × angular frequency.

Answer: (b) 0.134 m

Explanation:
1. Tension at the bottom: \( T_1=2g \); at the top: \( T_2=(2+8)g=10g \).
2. Speed \( v\propto\sqrt T \) and the frequency stays the same, so \( \lambda\propto\sqrt T \).
3. \( \lambda_2=0.06\sqrt{\frac{10g}{2g}}=0.06\sqrt5=0.134\ \text{m} \)

Teacher's Note:
1. Frequency is fixed by the source and does not change as the pulse travels.
2. Tension increases upwards because the rope supports its own weight.

Answer: (c) 100 Hz, 4 cm, 4 m/s

Explanation:
1. Mean position to the next mean position takes half a period: \( \frac{T}{2}=5\ \text{ms}\Rightarrow T=10\ \text{ms},\ f=100\ \text{Hz} \).
2. Consecutive particles at the mean position are half a wavelength apart: \( \frac{\lambda}{2}=2\ \text{cm}\Rightarrow\lambda=4\ \text{cm} \).
3. \( v=f\lambda=100\times0.04=4\ \text{m/s} \)

Teacher's Note:
1. A particle returns to its mean position twice per cycle.
2. Two consecutive points of zero displacement are λ/2 apart.

Answer: (d) 4

Explanation:
1. \( v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{T}{\rho A}} \), so \( v\propto\frac{1}{\sqrt\rho} \).
2. \( \frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\frac12\Rightarrow\frac{\rho_1}{\rho_2}=4 \)

Teacher's Note:
1. With the same tension and area, μ is proportional to the density.
2. Slower wave means denser wire.

Answer: (d) 0.5 W

Explanation:
1. \( \mu=\frac{T}{v^2}=\frac{100}{10^4}=0.01\ \text{kg/m} \).
2. \( P=2\pi^2\mu vf^2A^2=2(10)(0.01)(100)(100)(25\times10^{-6})=0.5\ \text{W} \)

Teacher's Note:
1. Get μ from T and v first.
2. Power = 2π²μ v f² A².

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

Explanation:
1. After reflection the wave travels in the −x direction, so x and vt appear with the same sign: \( vt+x \).
2. Reflection from a rigid boundary introduces a phase change of π, which gives a negative sign. The amplitude is 0.8a.

Teacher's Note:
1. Reflection at a rigid end inverts the wave.
2. The reflected wave travels backwards, so the sign of x changes.

Answer: (d) π and 0

Explanation:
1. \( \Delta\phi=\frac{2\pi}{\lambda}\Delta x=\frac{2\pi}{2}\times1=\pi \).
2. \( A=2A_0\cos\frac{\Delta\phi}{2}=0 \)

Teacher's Note:
1. Phase difference = (2π/λ) × path difference.
2. A phase difference of π gives destructive interference.

Answer: (c) 0.25 cm, 120 cm/s, 3 cm

Explanation:
1. A standing wave \( 2A\sin kx\cos\omega t \) has component waves of amplitude \( A=\frac{0.50}{2}=0.25\ \text{cm} \).
2. \( v=\frac{\omega}{k}=\frac{40\pi}{\pi/3}=120\ \text{cm/s} \).
3. Distance between nodes: \( \frac{\lambda}{2}=\frac{\pi}{k}=3\ \text{cm} \)

Teacher's Note:
1. The amplitude of each component is half the maximum amplitude of the standing wave.
2. Adjacent nodes are separated by π/k.

Answer: (d) 10 cm, 30 cm, 50 cm

Explanation:
1. Nodes occur where \( \cos\frac{\pi x}{20}=0 \), that is \( \frac{\pi x}{20}=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2} \).
2. So x = 10, 30 and 50 cm (within 60 cm). Antinodes are at x = 0, 20, 40 and 60 cm.

Teacher's Note:
1. Nodes are where the spatial factor vanishes.
2. The cosine form means antinodes (not nodes) sit at the ends of the interval x = 0.

Answer: (a) 0.5 m/s

Explanation:
1. The pulse peak moves from x = 0 to x = 1 m in 2 s.
2. \( v=\frac{1\ \text{m}}{2\ \text{s}}=0.5\ \text{m/s} \)

Teacher's Note:
1. Follow a fixed point of the shape, such as the peak.
2. Replacing x by x − vt shifts the shape by vt.

Answer: (b) 3.7 cm and 78 cm/s

Explanation:
1. Phase \( =0.56(1)+12(1)+\frac{\pi}{6}=13.084\ \text{rad} \).
2. \( y=7.5\sin(13.084)=7.5\times0.4945=3.7\ \text{cm} \).
3. \( v_y=A\beta\cos(13.084)=7.5\times12\times0.869=78\ \text{cm/s} \)

Teacher's Note:
1. Evaluate the sine and cosine in radians.
2. Transverse velocity is the time derivative of y.

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

Explanation:
1. \( k=\frac{2\pi}{\lambda}=2\pi\ \text{m}^{-1},\ \omega=2\pi f=4\pi\ \text{s}^{-1} \).
2. A wave moving in the −x direction has x and t with the same sign: \( \cos(2\pi x+4\pi t) \).

Teacher's Note:
1. Check k, ω and the relative sign of x and t.
2. Option (a) has the right k and ω but travels in the +x direction.

Answer: (d) 5.0 m/s

Explanation:
1. Opposite phase and within one cycle means separation = half a wavelength: \( \frac{\lambda}{2}=10\Rightarrow\lambda=20\ \text{m} \).
2. \( v=\frac{\lambda}{T}=\frac{20}{4}=5\ \text{m/s} \)

Teacher's Note:
1. One at crest and one at trough: path difference is an odd multiple of λ/2.
2. 'Within a single cycle' rules out 3λ/2 and above.

Answer: (b) \( \omega_0\text{ and }\frac{A_0}{\sqrt2} \)

Explanation:
1. The frequency is set by the source and cannot change along the string.
2. \( P=\tfrac12\mu\omega^2A^2v\propto A^2 \), so halving the power reduces the amplitude to \( \frac{A_0}{\sqrt2} \).

Teacher's Note:
1. Frequency of a wave is determined entirely by its source.
2. Power is proportional to the square of the amplitude at fixed frequency.

Answer: (a) The rate at which energy is carried is directly proportional to the velocity of the wave

Explanation:
1. \( P=\tfrac12\rho\,\omega^2y_m^2V \).
2. P is proportional to ω² and y_m², so doubling either gives four times the power; P is directly proportional to V for fixed ρ, ω and y_m.

Teacher's Note:
1. Write the power expression and test each statement.
2. Power depends on the square of frequency and amplitude.

Answer: (b) 30 Hz

Explanation:
1. \( v=\sqrt{\frac{100}{0.04}}=50\ \text{m/s} \).
2. \( P=2\pi^2\mu vf^2A^2\Rightarrow f^2=\frac{90}{2(10)(0.04)(50)(0.05)^2}=900 \), so f = 30 Hz.

Teacher's Note:
1. Compute the wave speed first.
2. Rearrange P = 2π²μvf²A² for f.

Answer: (c) v′ < v

Explanation:
1. Reflection with inversion occurs when the pulse meets a denser (higher linear density) string.
2. Wave speed is \( \sqrt{T/\mu} \), so the transmitted wave in the heavier string is slower: v′ < v.

Teacher's Note:
1. A pulse reflects with a phase change of π at a heavier medium.
2. The tension is the same on both sides.

Answer: (a) The maximum intensity that can be achieved at a point is twice the intensity of either wave and occurs if φ = 0

Explanation:
1. At φ = 0 the amplitude is \( 2y_m \), so the intensity (\( \propto A^2 \)) is four times that of either wave.
2. Therefore the statement that the maximum intensity is twice that of either wave is incorrect.

Teacher's Note:
1. Intensity is proportional to the square of the amplitude.
2. Constructive interference of equal waves doubles the amplitude and quadruples the intensity.

Answer: (a) 5

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

Teacher's Note:
1. Treat each wave as a vector whose direction is its phase.
2. The 10 and 7 μm waves are in opposite directions.

Answer: (b) 70 Hz

Explanation:
1. Successive resonant frequencies differ by the fundamental: \( 420-350=70\ \text{Hz} \).
2. Check: 350 = 5 × 70 and 420 = 6 × 70.

Teacher's Note:
1. For a string fixed at both ends, f_n = n f₁, so consecutive harmonics differ by f₁.
2. Confirm that both given frequencies are integer multiples of the result.

Answer: (d) 6/7

Explanation:
1. Stationary wave: \( \sin kx_1=\sin\frac{\pi}{3}>0 \) and \( \sin kx_2=\sin\frac{3\pi}{2}<0 \), so the particles are in opposite phase: \( \phi_1=\pi \).
2. Travelling wave: \( \phi_2=k(x_2-x_1)=\frac{3\pi}{2}-\frac{\pi}{3}=\frac{7\pi}{6} \).
3. \( \frac{\phi_1}{\phi_2}=\frac{\pi}{7\pi/6}=\frac67 \)

Teacher's Note:
1. In a stationary wave all points between two nodes oscillate in phase, and neighbouring segments are in opposite phase.
2. In a travelling wave the phase difference is k times the separation.

Answer: (a) 1 : 2

Explanation:
1. \( f=\frac{1}{2L}\sqrt{\frac{T}{\mu}},\ \mu=\rho\pi r^2 \).
2. \( \frac{f_A}{f_B}=\sqrt{\frac{T_A}{T_B}\cdot\frac{\rho_Br_B^2}{\rho_Ar_A^2}}=\sqrt{\frac{2}{2\times4}}=\frac12 \)

Teacher's Note:
1. Linear density is proportional to density times the square of the radius.
2. The length is the same, so it cancels.

Answer: (c) 40 cm

Explanation:
1. For a string fixed at both ends, \( \lambda_n=\frac{2L}{n} \).
2. \( \lambda_4=\frac{2\times80}{4}=40\ \text{cm} \) (the tension and mass do not matter for the wavelength).

Teacher's Note:
1. The wavelength of a harmonic depends only on the length of the string and the harmonic number.
2. The speed and frequency depend on T and μ, but the wavelength does not.

Answer: (c) \( \frac{a}{2} \)

Explanation:
1. \( a\sin\omega t\cos kx=\frac a2\left[\sin(\omega t+kx)+\sin(\omega t-kx)\right] \).

Teacher's Note:
1. Use the product-to-sum identity.
2. The maximum amplitude of the standing wave, a, is twice that of each component.