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Read MoreWelcome to the NCERT Solutions for Class 11 Physics - Chapter Waves. This page offers a step-by-step solution to the specific question from Exercise 1, Question 8: . With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.
A transverse harmonic wave on a string is described by
y(x,t) = 3.0 sin [36t + 0.018x + π /4]
Where x and y are in cm and t in s. The positive direction of x is from left to right.
(a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
(b) What are its amplitude and frequency?
(c) What is the initial phase at the origin?
(d) What is the least distance between two successive crests in the wave?
(a) Yes; Speed = 20 m/s, Direction = Right to left
(b) 3 cm; 5.73 Hz
(c) π /4
(d) 3.49 m
Explanation:
Given,
y(x, t) =3 sin (36t +0.018x + π/4) . . . . . . . . . . . ( 1 )
( i ) We know, the equation of a progressive wave travelling from right to left is:
y (x, t) = a sin (ωt + kx + Φ) . . . . . . . . . . . . . . . . . . . ( 2 )
Comparing equation ( 1 ) to equation ( 2 ), we see that it represents a wave travelling from right to left and also we get:
a = 3 cm, ω = 36 rad/s , k = 0.018 cm and ϕ = π/4
( ii )Therefore the speed of propagation , v = ω/k = 36/ 0.018 = 20 m/s
( iii ) Amplitude of the wave, a = 3 cm
Frequency of the wave v = ω / 2π = 36 /2π = 5.7 hz
( iv ) Initial phase at the origin = π/4
( v ) the smallest distance between two adjacent crests in the wave, λ = 2π/ k = 2π / 0.018 = 349 cm
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Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 8: A transverse harmonic wave on a string is described by y(x,t) = 3.0 sin [36t + 0.018x....
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