#### The Transverse Wave

A sinusoidal transverse wave is traveling along a string in the negative direction of an *x* axis. The figure below shows a plot of the displacement as a function of position at time *t* = 0. The *x* axis is marked in increments of 20 cm and the *y* axis is marked in increments of 0.5 cm. The string tension is 4.4 N, and its linear density is 25 g/m.

(a) Find the amplitude. m

(b) Find the wavelength. m

(c) Find the wave speed. m/s

(d) Find the period of the wave. s

(e) Find the maximum speed of a particle in the string. m/s

(f) Complete the equation describing the traveling sinusoidal transverse wave, in which *x* and *y* are in meters and *t* is in seconds.

*y*(*x*, *t*) = sin( ?*x* + ?*t* + ? )

#### Answer

The amplitude of the sinusoidal transverse wave is the maximum elongation from the x axis: $A = 5*0.5 cm = 2.5 cm$

and its wavelength is the distance between two consecutive maxima $\lambda = 4*20 cm =80 cm$

and the wave speed is $v = \sqrt{T/\mu} =13.266 m/s$

The equation that relates the wavelength to the wave speed is:

$\lambda = v*T$ and thus

$T = 0.0603 seconds$ and

$V max = \omega*A =(2*\pi/T)*A =(2*\pi/0.0603)*0.025= 2.605 m/s$

The equation of the wave traveling in positive x direction is

$Y(x,t) = A*sin[2*\pi*(x/\lambda -t/T) +\phi]$

The equation of wave travelling in the NEGATIVE x direction is

$Y(x,t) = A*sin(2*\pi*(-x/\lambda -t/T) +\phi) = A*sin[2*\pi(x/\lambda +t/T) +\phi]=$ $=0.025*sin(7.85*x +104.2*t +\phi) meter$

From the initial conditions at $t=0$ (from the picture) we can determine the phase $\phi$

$0.02 =0.025*sin(\phi)$ $\phi = 53.13 deg =0.927 rad$

Therefore the equation of this wave is

$Y(x,t)= 0.025*sin(7.85x +104.2t +0.927)$

You can continue reading a discussion on transversal waves at physics forums