SATHEE: Waves - Result Question 11

Waves - Result Question 11

11. The phase difference between two waves, represented by

$y_{1} = 10^{- 6} \sin 100 t + (x / 50) + 0.5 m$ $y_{2} = 10^{- 6} \cos 100 t + (x / 50) m$ where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately [2004]

(a) 1.5 radians

(b) 1.07 radians

(d) 0.5 radians

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Answer:

Correct Answer: 11. (b)

Solution:

(b) $y_{1} = 10^{- 6} \sin (100 t + x / 50 + 0.5) m$

$= 10^{- 6} \cos (100 t + x / 50 - π / 2 +$

$0.5) m$

$\begin{aligned} y_{2} = 10^{- 6} \cos (100 t + x / 50) m \\ ∴ ϕ = π / 2 - 0.5 = 1.07 r a d \end{aligned}$

Waves - Result Question 11

11. The phase difference between two waves, represented by

Answer:

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Waves - Result Question 11

11. The phase difference between two waves, represented by

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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