91Ó°ÊÓ

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

$y(x,t)=A\cos (kx∓\mathrm{Ó\neg }t)$--(1)

Where A, k and$\mathrm{Ó\neg }$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$\mathrm{λ}$:

$\mathrm{λ}=\frac{2\mathrm{π}}{k}$--(2)

The relation between the wave speed v and the wave angular speed$\mathrm{Ó\neg }$is :

$v=\frac{\mathrm{Ó\neg }}{k}$--(3)

And the wave speed in a string in terms of the tension T and the linear mass density$\mathrm{μ}$:

$V=\sqrt{\frac{T}{\mathrm{μ}}}$--(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:

$A=8.50×{10}^{-3}m$

The wave number is:

$k=172m^{-1}$

The angular speed is:

$\mathrm{Ó\neg }=4830s^{-1}$

Put in the values for$\mathrm{Ó\neg }$ 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:

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

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

$\begin{array}{r}\mathrm{μ}=\frac{m}{l}\\ =\frac{1.28×{10}^{−3}\mathrm{kg}}{1.50\mathrm{m}}\\ =8.50×{10}^{−4}\mathrm{kg}/\mathrm{m}\end{array}$

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$\mathrm{μ}$into equation (4) and solve for W:

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

$W=0.671\mathrm{N}$

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

$\mathrm{λ}=\frac{2\mathrm{π}}{172m^{-1}}=0.037m$

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

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

Now, put in the values for $\mathrm{λ}$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 $y(x,t)=(8.50mm)cos(172m^{-1}x+4830s^{-1}t)$.