Question

One end of a long string of linear mass density $8.0\times {10}^{-3}$ 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 1

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, $\mu =8.0\times {10}^{-3}kg{m}^{-1}$
Frequency of the tuning fork, ν = 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, $T=mg=90\times 9.8=882N$
The velocity of the transverse wave v, is given by the relation:
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{882}{8.0\times {10}^{-3}}}=332m/s$
Angular frequency, $\omega =2\pi v$
$=2\times 3.14\times 256=1608.5=1.6\times {10}^{3}rad/s.....\left(iii\right)$
Wavelength, $\lambda =\frac{v}{v}=\frac{332}{256}m$
$\therefore$ Propagation constant, $k=\frac{2\pi }{\lambda }$
$=\frac{2\times 3.14}{\frac{332}{256}}=4.84m^{-1}....\left(iv\right)$
Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:
$y(x,t)=0.05sin(1.6\times {10}^{3}t-4.84x)m$