Question

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is $3.5\times {{10}^{-2}}kg$and its linear mass density is$4.0\times {{10}^{-2}}kg{{m}^{-1}}$. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

Answer 1

The speed of a transverse wave on the string is the frequency of the vibration times the wavelength of the vibration. The tension in the string is the product of the square of the speed of a transverse wave on the string and the linear mass density.
Formula used:

& v=\upsilon \lambda \\\ & T={{v}^{2}}\mu \\\ \end{aligned}$$ **Complete answer:** From the given information, we have the data as follows. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is $$3.5\times {{10}^{-2}}kg$$and its linear mass density is$$4.0\times {{10}^{-2}}kg{{m}^{-1}}$$. The mass of the wire, $$m=3.5\times {{10}^{-2}}kg$$ The linear mass density is, $$\mu =4.0\times {{10}^{-2}}kg{{m}^{-1}}$$ The frequency of vibration is, $$\upsilon =45\,Hz$$ The length of the wire is computed as mass of the wire by the linear mass density. Thus, the length of the wire is, $$l=\dfrac{m}{\mu }$$ Substitute the values in the above equation. $$\begin{aligned} & l=\dfrac{3.5\times {{10}^{-2}}}{4.0\times {{10}^{-2}}} \\\ & \therefore l=0.875 \\\ \end{aligned}$$ The wavelength of the stationary wave is twice the length of the wire by the number of nodes in the wire. So, we have, $$\lambda =\dfrac{2l}{n}$$ Substitute the values in the above equation. $$\begin{aligned} & \lambda =\dfrac{2\times 0.875}{1} \\\ & \therefore \lambda =1.75\,m \\\ \end{aligned}$$ The speed of the transverse wave in the string is given as follows. $$v=\upsilon \lambda $$ Substitute the values in the above formula $$\begin{aligned} & v=45\times 1.75 \\\ & \therefore v=78.75m{{s}^{-1}} \\\ \end{aligned}$$ The tension produced in the string is given by the formula as follows. $$T={{v}^{2}}\mu $$ Where $$v$$ is the speed of the transverse wave and $$\mu $$ is the linear mass density. Substitute the values in the above formula. $$\begin{aligned} & T={{78.75}^{2}}\times 4.0\times {{10}^{-2}} \\\ & \therefore T=248.06\,N \\\ \end{aligned}$$ **$$\therefore $$ (a) the speed of a transverse wave on the string is $$78.75m{{s}^{-1}}$$, and (b) the tension in the string is 248.06 N.** **Note:** The length of the wire is computed as a mass of the wire by the linear mass density. The wavelength of the stationary wave is twice the length of the wire by the number of nodes in the wire.