Question

Suppose that 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 having a frequency $256Hz$. The other end passes over a pulley and is tied to a pan having a mass of $90kg$. The pulley end takes up all the incoming energy. Because of that the reflected waves at this end will have negotiable amplitude. At $t=0$, the left end which is the fork end of the string, $x=0$ has no transverse displacement, that is $y=0$, and is moving along positive y-direction. The wave amplitude will be $5cm$. What will be the transverse displacement $\left( y \right)$ function of $x$ and $t$ which explains the wave on the string?

Answer 1

first of all find the tension on the string. Calculate the velocity of the transverse wave. And also find the angular frequency and wavelength of the wave. Using this determines the propagation constant. Substitute all these values in the general equation of displacement.

Complete answer:
The equation of a travelling wave which is propagating along the positive y-direction can be written as,
$y\left( x,t \right)=a\sin \left( \omega t-kx \right)$
Linear mass density is given as,
$\mu =8\times {{10}^{-3}}kg{{m}^{-1}}$
Frequency of the tuning fork will be,
$\nu =256Hz$
Amplitude of the wave can be written as,
$a=0.05m$
Mass of the pan is,
$m=90kg$
The tension in the string will be found as,
$T=mg=90\times 9.8=882N$
The velocity of the transverse can be found as,
$v=\sqrt{\dfrac{T}{\mu }}$
Substituting the values in it,
$\Rightarrow v=\sqrt{\dfrac{882}{8\times {{10}^{-3}}}}=332m{{s}^{-1}}$
The angular frequency will be found as,
$\Rightarrow \omega =2\pi \nu =2\pi \times 56=160.85=1.6\times {{10}^{3}}rad{{s}^{-1}}$
The wavelength of the wave,
$\Rightarrow \lambda =\dfrac{v}{\nu }=\dfrac{332}{256}m$
Propagation constant can be written as,
$\Rightarrow k=\dfrac{2\pi }{\lambda }=\dfrac{2\pi }{\dfrac{332}{256}}=4.84{{m}^{-1}}$
Substituting the values in the general equation gives,
$\Rightarrow y\left( x,t \right)=0.05\sin \left( 1.6\times {{10}^{3}}t-4.84x \right)m$

Note:
A transverse wave is defined as the wave which is having the direction of propagation of the wave perpendicular to the vibration of the matter in it. The ripples produced in the surface of water and also all the electromagnetic waves like light waves, microwaves and so on are the example for the transverse wave.