Principle of Superposition of Waves and Reflection of Waves NEET Questions

Physics - Oscillations and Waves

Principle of Superposition of Waves and Reflection of Waves MCQ Questions


7.
For two waves with equal amplitude a and phase difference φ, complete destructive interference (zero resultant amplitude) occurs when:
A.
φ = 2π giving A = 0
B.
φ = π (or 3π, 5π, …) giving A = 2a cos(π/2) = 0
C.
φ = π/2 giving A = 0
D.
φ = 0 giving A = 0
😑 View Answer Rough Work Error
ANSWER :
B. φ = π (or 3π, 5π, …) giving A = 2a cos(π/2) = 0
8.
Two waves y₁ = a sin(kx−ωt) and y₂ = a sin(kx−ωt+π) are superposed. The resultant y = y₁ + y₂ equals:
A.
y = 0 everywhere at all times (complete destructive interference)
B.
y = 2a sin(kx−ωt+π/2)
C.
y = a sin(2kx−2ωt)
D.
y = 2a sin(kx−ωt)
😑 View Answer Rough Work Error
ANSWER :
A. y = 0 everywhere at all times (complete destructive interference)
9.
Two waves with equal amplitude a are in phase (φ = 0). The resultant displacement and intensity compared to a single wave are:
A.
Displacement amplitude = a; Intensity = same as single wave
B.
Displacement amplitude = 2a; Intensity = 2 × single wave intensity
C.
Displacement amplitude = a√2; Intensity = 2 × single wave intensity
D.
Displacement amplitude = 2a (doubled); Intensity = 4 × single wave intensity
😑 View Answer Rough Work Error
ANSWER :
D. Displacement amplitude = 2a (doubled); Intensity = 4 × single wave intensity
10.
The resultant of two waves y₁ = a sin(kx−ωt) and y₂ = a cos(kx−ωt) is:
A.
y = 2a sin(kx−ωt) with amplitude 2a
B.
y = a√2 sin(kx−ωt+π/4) with amplitude a√2 and phase shift π/4
C.
y = a sin(kx−ωt+π/2) with amplitude a
D.
y = 0 since sine and cosine cancel
😑 View Answer Rough Work Error
ANSWER :
B. y = a√2 sin(kx−ωt+π/4) with amplitude a√2 and phase shift π/4
11.
For two superposing waves A sin(kx−ωt) and B cos(kx−ωt), the resultant amplitude is:
A.
√(A² − B²)
B.
A − B
C.
√(A² + B²)
D.
A + B
😑 View Answer Rough Work Error
ANSWER :
C. √(A² + B²)
12.
Two sinusoidal waves of unequal amplitudes a₁ and a₂ and phase difference φ are superposed. The resultant amplitude is:
A.
√(a₁² + a₂² + 2a₁a₂ cosφ)
B.
a₁ + a₂
C.
a₁ − a₂
D.
√(a₁² − a₂² + 2a₁a₂ sinφ)
😑 View Answer Rough Work Error
ANSWER :
A. √(a₁² + a₂² + 2a₁a₂ cosφ)