18 A 1.50m string of weight 0.0125N is tied to the ceiling at its upper end, and the lower end supports a weight W. Ignore the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation

y (x, t) = (8.50 mm) cos (172 rad/mx 4830 rad/s t)

Assume that the tension of the string is constant and equal to W. (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight W? (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string?

The wave function y (x, t) of a sinusoidal wave which describes the displacement of individual particles in the medium is:

$y(x,t)=A\cos (kx\mp \omega t)$--(1)

Where A, k and$\omega$are constants.

The minus sign is used when the wave is traveling in positive X-direction and the plus sign is used when the wave is traveling in the negative X-direction.

The relation between the wave number k and the wavelength$\lambda$:

$\lambda =\frac{2\mathrm{\pi }}{k}$--(2)

The relation between the wave speed v and the wave angular speed$\omega$is :

$v=\frac{\omega }{k}$--(3)

And the wave speed in a string in terms of the tension T and the linear mass density$\mu$:

$V=\sqrt{\frac{T}{\mu }}$--(4)

The length of the string is l = 1.50m, its weight is T_g = 0.0125N, the tension in the string is T = W and the wave function of the wave traveling in the string is:

$y(x,t)=(8.50mm)cos(172m^{-1}x-4830s^{-1}t)$

Compare the wave function of the string with the wave function in equation (1),
The wave amplitude is:

$A=8.50\times {10}^{-3}m$

The wave number is:

$k=172m^{-1}$

The angular speed is:

$\omega =4830s^{-1}$

Put in the values for$\omega$ and k into equation (3):

$\begin{array}{l}v=\sqrt{\frac{4830s^{-1}}{172m^{-1}}}\\ =28.1mls\end{array}$

The mass of the string is:

$\begin{array}{r}m=\frac{F_g}{g}\\ =\frac{0.0125\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\\ =1.28\times {10}^{-3}\mathrm{kg}\end{array}$

The linear mass density is the mass length, so the linear of the string is:

$\begin{array}{r}\mu =\frac{m}{l}\\ =\frac{1.28\times {10}^{-3}\mathrm{kg}}{1.50\mathrm{m}}\\ =8.50\times {10}^{-4}\mathrm{kg}/\mathrm{m}\end{array}$

The speed of any motion in terms of the distance travelled x and the time interval t is:

$$\begin{gathered} v=\frac{x}{t} \\ t=\frac{x}{v} \end{gathered}$$

Substitute v and x = l, therefore the time it takes for the wave to travel the full length of the string:

role="math" localid="1668150737173" $$\begin{gathered} \begin{array}{l}t=\frac{1.50m}{28.1mls}\\ =0.053s\end{array} \\ t=0.053s \end{gathered}$$

Put in the values for v, T and$\mu$into equation (4) and solve for W:

$\begin{array}{r}28.08\mathrm{m}/\mathrm{s}=\sqrt{\frac{W}{8.50\times {10}^{-4}\mathrm{kg}/m}}\\ \begin{array}{rl}W& =(28.08{)}^2\times \left(8.50\times {10}^{-4}\right)\\ & =0.671\mathrm{N}\end{array}\\ \end{array}$

$W=0.671\mathrm{N}$

Put in the value for k into equation (2), so we get the wavelength W:

$\lambda =\frac{2\mathrm{\pi }}{172m^{-1}}=0.037m$

The number of wavelengths on the string at any instant of time can be calculated from the following relations:

$$\begin{gathered} n=\frac{Thelengthofthestring}{Thewavelength} \\ =\frac{l}{\lambda } \end{gathered}$$

Now, put in the values for $\lambda$and l:

$\begin{array}{c}n=\frac{1.50}{0.037m}\\ =411\end{array}$

n = 41.1wavelengths

The minus sign in the cosine function for the upward direction means that the positive direction is upward.

If the wave is traveling downward (the negative direction), then same wave equation is used but with plus sign inside the cosine function:

$y(x,t)=(8.50mm)cos(172m^{-1}x+4830s^{-1}t)$

Therefore, the time taken for a pulse to travel the full length of the string is 0.053s, the weight (W) is 0.671N, the number of wavelengths on the string at any instant of time is 41.1 and the equation of motion for waves travelling down the string is $y(x,t)=(8.50mm)cos(172m^{-1}x+4830s^{-1}t)$.