91Ó°ÊÓ

• Equation of a transverse wave on a string is- $y=\left(2.0mm\right)\sin \left[\left(20m^{-1}\right)x-\left(600s^{-1}\right)t\right]$
• Tension in the string, T = 15 N
• Angular frequency of the wave, $\mathrm{Ó\neg }=600s^{-1}$
• Wavelength of the wave, $\mathrm{λ}=0.05m$

The product of the frequency and wavelength of the wave, gives us the wave speed. The linear density is the term used in place of the mass per unit length of the string.

Formula:

The speed of the wave,$v=n×\mathrm{λ}$ (i)

The formula for the linear density of distribution,$d=\frac{T}{v^2}$ (ii)

Using the equation (i) and the required given values, the wave speed is given as:

$$\begin{gathered} v=600s^{-1}×0.05m \\ =30m/s \end{gathered}$$

Hence, the value of the wave speed is

Using equation (i) and the value of wave speed, we get the linear density as:

$$\begin{gathered} d=\frac{15N}{{\left(30m/s\right)}^2} \\ =1.6667×{10}^{-2}kg/m \\ ≈1.7×{10}^{-2}kg/m \end{gathered}$$

Hence, the value of the linear density of the string is$1.7×{10}^{-2}kg/m$