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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 22:
A travelling harmonic wave on a string is described by
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
. 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 travelling harmonic wave on a string is described by
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?
(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.
(a) The given harmonic wave is:
y(x,t) = 7.5sin [0.0050x + 12t + π/4]
For x = 1 cm and t = 1s,
y = (1, 1) = 7.5sin [0.0050 + 12 + π/4]
.= 7.5sin [12.0050 + π/4]
= 7.5 sinθ
Where, θ = 12.0050 + π/4 = 12.0050 + 3.14 / 4 = 12.79 rad
= 180 /3.14 x 12.79 = 732.81°
∴ y = (1, 1) = 7.5sin [732.81°]
= 7.5 sin (90 x 8 + 12.81°)
= 7.5 sin (12.81°)
= 7.5 x 0.2217
= 1.6629 ≈ 1.663 cm
The velocity of the oscillation at a given point and time is given as:
v = d/dt y(x,t) = d/dt [7.5sin(0.0050x + 12t +π/4)]
= 7.5 x 12cos (0.0050x + 12t +π/4)
At x = 1 cm and t = 1s:
v = y(1,1) = 90 cos (12.005 + π/4)
= 90cos(732.81°) = 90cos(90 x 8 + 12.81°)
= 90cos(12.81°)
= 90 x 0.975 = 87.75 cm/s
Now, the equation of a propagating wave is given by:
y(x,t) = a sin(kx + wt + ø)
Where,
k = 2π / λ
∴ λ = 2π / k
And ω = 2πv
∴ v = ω / 2π
Speed = v = vλ = ω / k
Where
ω = 12 rad/s
k = 0.0050 m-1
∴ v = 12 /0.0050 = 2400 cm/s
∴ Hence, the velocity of the wave oscillation at x = 1 cm and t = 1 s is not equal to the velocity of the wave propagation.
(b) Propagation constant is related to wavelength as:
k = 2π / λ
∴ λ = 2π / k = 2 x 3.14 / 0.0050
= 1256 cm = 12.56
Therefore, all the points at distances nλ , (n =±1, ±2....and so on) i.e. ± 12.56 m, ± 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.
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 22: A travelling harmonic wave on a string is described by y(x,t) = 7.5sin [0.0050x + 12t + π/4] ....
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