Sinusoidal transverse wave

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.

Sinusoidal Transverse Wave

(a) Find the amplitude.  m
(b) Find the wavelength. 
(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(xt) =  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


valentin68

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