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 positivey-direction is given by the displacement equation:

$y(x,t)=asin(wt-kx)$ ...(i)
Linear mass density, $\mu =8.0\times 10-3kg{m}^{-1}$

Frequency of the tuning fork, $v=256Hz$

Amplitude of the wave, $a=5.0cm=0.05m$ ...(ii)

Mass of the pan, m $=90kg$
Tension in the string, $T=mg=909\times 9.8=882N$

The velocity of the transverse wave v, is given by the relation:

$v=\sqrt{\frac{T}{\mu }}$

$=\frac{882}{8.0\times {10}^{-3}}=332m/s$

Angular Frequency, $\omega =2\pi f$

$=2\times 3.14\times 256$

$=1608.5=1.6\times {10}^{3}rad/s$ ...(iii)

Wavelength, $\lambda =\frac{v}{f}=\frac{332}{256}m$

$\therefore$ Propagation constant, $k=\frac{2\pi }{\lambda }$

$=\frac{2\times 3.14}{\frac{332}{256}}=4.84m^{-1}$ ...(iv)

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get the displacement equation:

$y(x,t)=0.05sin(1.6\times {10}^{3}t-4.84x)m$.