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Academic session 2026-27, Chapter 14 in the current curriculum.
Welcome 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 11:
The transverse displacement of a wire (clamped at both its ends) is described as :
y (x, t) = 0.06 sin(2π/3x) cos(120πt)
The mass of the wire is 6 x 10-2 kg and its length is 3m. Provide answers to the following questions:
( I ) Is the function describing a stationary wave or a travelling wave?
( ii ) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave.
( iii ) Calculate the wire’s tension. [X and y are in meters and t in secs]
. 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.The transverse displacement of a wire (clamped at both its ends) is described as :
y (x, t) = 0.06 sin(2π/3x) cos(120πt)
The mass of the wire is 6 x 10-2 kg and its length is 3m. Provide answers to the following questions:
( I ) Is the function describing a stationary wave or a travelling wave?
( ii ) Interpret the wave as a superposition of two waves travelling in opposite directions. Find the speed, wavelength and frequency of each wave.
( iii ) Calculate the wire’s tension. [X and y are in meters and t in secs]
We know,
The standard equation of a stationary wave is described as:
y (x, t) = 2a sin kx cos ωt
Our given equation y (x, t) =0.06sin(2π/3x)cos(120πt) is similar to the general equation .
( i ) Thus, the given function describes a stationary wave.
( ii ) We know, a wave travelling in the positive x – direction can be represented as :
y1 = a sin( ωt – kx )
Also,
Wave travelling in the negative x – direction is represented as :
y2 = a sin( ωt + kx )
Super- positioning these two waves gives us :
y = y1 + y2
= a sin( ωt – kx ) – a sin( ωt + kx )
= asin(ωt)cos(kx) – asin(kx)cos(ωt) – asin(ωt)cos(kx) – asin(kx)cos(ωt)
= – 2asin(kx)cos(ωt)
= −2asin(2πλx)cos(2πvt) . . . . . . . . . . . . . . ( 1 )
The transverse displacement of the wires is described as :
0.06sin(2π3x)cos(120πt) . . . . . . . . . . . . . . . . .( 2 )
Comparing equations ( 1 ) and ( 2 ) , we get :
2π/ λ = 2π/ 3
Therefore, wavelength λ = 3m
Also, 2πv/λ = 120π
Therefore, speed v = 180 m/s
And, Frequency = v/λ = 180/3
= 60 Hz
( iii )Given, Velocity of the transverse wave, v = 180 m / s
The string’s mass, m = 6 × 10-2 kg
String length, l = 3 m
Mass per unit length of the string, μ = m/l = (6 x 10-2 )/3
= 2 x 10-2 kg/m
Let the tension in the wire be T
Therefore, T = v2 μ
= 1802 x 2 x 10-2
= 648 N.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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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 11: The transverse displacement of a wire (clamped at both its ends) is described as : y (x, t) = 0.0....
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