The following four waves are sent along strings with the same linear densities (xis in meters and tis in seconds). Rank the waves according to (a) their wave speed and (b) the tension in the strings along which they travel, greatest first:

(1)${\mathit{Y}}^{\mathbf{1}}\mathbf{=}\mathbf{\left(}\mathbf{3}\mathit{m}\mathit{m}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{3}\mathit{t}\mathbf{)}\mathbf{,}$ (3)${\mathit{y}}^{\mathbf{3}}\mathbf{=}\mathbf{\left(}\mathbf{1}\mathit{m}\mathit{m}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{4}\mathit{x}\mathbf{-}\mathit{t}\mathbf{)}$,

(2) ${\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{\left(}\mathbf{6}\mathit{m}\mathit{m}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{-}\mathit{t}\mathbf{)}$, (4)${\mathit{y}}^{\mathbf{4}}\mathbf{=}\mathbf{\left(}\mathbf{2}\mathit{m}\mathit{m}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{2}\mathit{t}\mathbf{)}$.

The four waves along the strings with the same linear densities are,

$$\begin{gathered} Y_1=3mm)\sin (x-3t) \\ {y}_2=(6mm)\sin (2x-t) \\ {y}_3=(1mm)\sin (4x-t) \\ {y}_4=(2mm)\sin (x-2t) \end{gathered}$$

Use the concept of the equation of transverse wave and speed of a travelling wave. The wave speed on a stretched string gives the relation between speed and tension in the string.

Formulae are as follow:

$$\begin{gathered} y=y_m\sin (kx-\omega t) \\ v=\frac{\omega }{k} \\ v=\sqrt{\frac{T}{\mu }} \end{gathered}$$

Where, v is wave speed, T is tension in string, 𝝁 is mass per unit length, 𝝎 is angular frequency, k is wave number, t is time.

(a)

Rank the waves according to their wave speed :

The equation of transverse wave is,

$y=y_m\sin (kx-\omega t)...........................................................................\left(1\right)$

The speed of the travelling wave is,

$v=\frac{w}{k}$

The equation (i) is,

$y_1=\left(3mm\right)\sin (x-3t)$

Compare this equation with equation (1), then the speed of the travelling wave is,

$$\begin{gathered} v_1=\frac{3}{1} \\ {v}_1=3 \end{gathered}$$

The equation (ii) is,

$y_2=\left(6mm\right)\sin (2x-t)$

Compare this equation with equation (1), then the speed of the travelling wave is,

$$\begin{gathered} V_2=\frac{1}{2} \\ {v}_2=0.5 \end{gathered}$$

The equation (iii) is,

$y_3=\left(1mm\right)\sin 94x-t)$

Compare this equation with equation (1), then the speed of the travelling wave is,

$$\begin{gathered} V_3=\frac{1}{4} \\ {V}_3=0.25 \end{gathered}$$

The equation (iv) is,

$y_4=\left(2mm\right)\sin (x-2t)$

Compare this equation with equation (1), then the speed of the travelling wave is

$$\begin{gathered} V_4=\frac{2}{1} \\ {V}_4=2 \end{gathered}$$

Hence, the rank of the waves according to the wave speed is $V_1>V_4>V_2>V_3$ (greatest first).

(b)

Rank the waves according to tension:

The wave speed on a stretched string is,

$$\begin{gathered} V=\sqrt{\frac{T}{\mu }} \\ V\propto \sqrt{T} \end{gathered}$$

The speed on the stretched string is directly proportional to the tension in the string with the same linear density.

The speed on the stretched string for equation (i) is,

$v_1\propto \sqrt{T_1}$

The speed on the stretched string for equation (ii) is,

$V_2\propto \sqrt{T_2}$

The speed on the stretched string for equation (iii) is,

$V_3\propto \sqrt{T_3}$

The speed on the stretched string for equation (i) is,

$V_4\propto \sqrt{T_4}$

Hence, the rank of the waves according to their tension is $T_1>T_4>T_2>T_3$ (greatest first).

Therefore, the wave speed by using its expression and rank their values can be found, by using the expression of the speed on the stretched string, and can also find the tension in each string and rank their values.