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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 24:
One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
. 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.One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.
The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:
y (x, t) = a sin (wt - kx) … (i)
Linear mass density, μ = 8.0 x 10-3 kg m-1
Frequency of the tuning fork, v = 256 Hz
Amplitude of the wave, a = 5.0 cm = 0.05 m … (ii)
Mass of the pan, m = 90 kg
Tension in the string, T = mg = 90 × 9.8 = 882 N
The velocity of the transverse wave v, is given by the relation:
v = underoot T / μ
= underoot 882 / 8.0 x 10-3 = 332 m/s
Angular Frequency, ω = 2πv
= 2 x 3.14 x 256
= 1608.5 = 1.6 x 103 rad/s .....(iii)
Wavelength, λ = v / V = 332 / 256 m
∴ Propagation constant, k = 2π / λ
= 2 x 3.14 / 332/256 = 4.84 m-1 ......... (iv)
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
y (x, t) = 0.05 sin (1.6 × 103t - 4.84 x) m
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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 24: One end of a long string of linear mass density 8.0 x 10-3 kg m-1 is connected to an electrically dr....
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