Question

The linear density of a string is$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{k}\mathit{g}\mathbf{/}\mathit{m}$. A transverse wave on the string is described by the equation$\mathit{y}\mathbf{=}\left(0.021m\right)\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left[\left(2.0m^{-1}\right)x+\left(30s^{-1}\right)t\right]$. (a)What are the wave speed and (b) What is the tension in the string?

Short answer

a) The speed of the wave is 15m/s

b) The tension in the string is $3.60\times {10}^{-3}\mathrm{N}$.

Step-by-step solution

The given data

• The wave equation is given as$y=\left(0.021m\right)\sin \left[\left(2.0m^{-1}\right)x+\left(30s^{-1}\right)t\right]$
• Linear density of a string,$\left(\mathrm{\mu }\right)=1.6\times {10}^{-4}\mathrm{kg}/\mathrm{m}$
• Wavelength,$\left(\mathrm{\lambda }\right)=0.5\mathrm{m}$
• Frequency,$\left(f\right)=30s^{-1}$

Understanding the concept of wave equation

The product of wavelength and frequency of the wave is called speed of the wave. the speed of the wave in a stretched string is directly proportional to the square-root of the tension force and inversely proportional to the square-root of linear density of the string.

Formula:

The wave speed of the wave, $v=n\times \lambda$ (i)

The velocity of the wave,$v=\sqrt{\frac{T}{\mu }}$ (ii)

a) Calculation of the wave speed

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

$$\begin{gathered} \mathrm{v}=30{\mathrm{s}}^{-1}\times 0.5\mathrm{m} \\ =15\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of wave speed is 15 m/s

b) Calculation of tension in the string

Using equation (ii), the tension in the string is given as:

$$\begin{gathered} T=v^2\mu \\ ={\left(15\mathrm{m}/\mathrm{s}\right)}^2\times \left(1.6\times {10}^{-4}\mathrm{kg}/\mathrm{m}\right) \\ =3.60\times {10}^{-3}\mathrm{N} \end{gathered}$$

Hence, the value of the tension in the string is $3.60\times {10}^{-3}\mathrm{N}$.