Class 11 Physics - Chapter Waves NCERT Solutions | One end of a long string of linear mass

Welcome to the NCERT Solutions for Class 11th 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....
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.

Answer

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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