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