Speed of a Transverse Wave

a) Write an expression for the velocity in each case with different tension variables. Divide them, replace the conditions for the velocities to find the relations for the tensions.

b) Same hint as in a).

a) The velocity for a propagating transverse wave is:

\begin{equation*}
v=\sqrt{\frac{T}{\mu}}.
\end{equation*}

For the second case, the velocity can be written as:

\begin{equation*}
v'=\sqrt{\frac{T'}{\mu}}.
\end{equation*}

Dividing both equations and solving for \(T'\) we get:

\begin{equation*}
T'=T\left(\frac{v'}{v}\right)^2.
\end{equation*}

Since \(v' = 2v\), then:

\begin{equation*}
T'=4T,
\end{equation*}

or:

\begin{equation*}
T'=280 \, \text{N}.
\end{equation*}

b) Dividing the equations as in a), but solving for the velocity \(v'\) we obtain:

\begin{equation*}
v'=v\sqrt{\frac{T'}{T}}.
\end{equation*}

Since \(T' = 3T\), then:

\begin{equation*}
v' = v \sqrt{3},
\end{equation*}

or:

\begin{equation*}
v' \approx 86.6 \, \text{m/s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

(a) We need to find the tension required for the speed of the transverse waves to double. The speed \(v\) of transverse waves on a uniform string with tension \(T\), and mass linear density \(\mu\) is given by the following equation

\begin{equation}
v=\sqrt{\frac{T}{\mu}}.
\end{equation}

Elevating both sides of the equation above to the square we obtain

\begin{equation}
\label{vel}
v^2=\frac{T}{\mu}.
\end{equation}

For any other speed \(v'\) on the same string we can then write

\begin{equation}
\label{velp}
v'^2=\frac{T'}{\mu},
\end{equation}

where \(T'\) is the new tension and \(mu\) is still the same as in equation \eqref{vel} because neither the mass of the string, its length or their ratio change, it is the same string. We can then take the ration of equation \eqref{velp} and \eqref{vel} to obtain

\begin{equation}
\frac{v'^2}{v^2}=\frac{\frac{T'}{\mu}}{\frac{T}{\mu}},
\end{equation}

where after simplification it becomes

\begin{equation}
\label{ratios}
\left(\frac{v'}{v}\right)^2=\frac{T'}{T}.
\end{equation}

We can solve for \(T\) from the equation above to get

\begin{equation}
T'=T\left(\frac{v'}{v}\right)^2.
\end{equation}

Because we want the tension necessary for the velocity to double, then \(v'=2v\) thus

\begin{equation}
T'=T\left(\frac{2v}{v}\right)^2,
\end{equation}

or simplifying

\begin{equation}
T'=4T.
\end{equation}

Using the numerical values

\begin{equation}
T'=4(70\,\text{N}),
\end{equation}

\begin{equation}
T'=280\,\text{N}.
\end{equation}

(b) For the final part of the problem, we are asked to find the speed of the transverse waves if the tension on the cord is tripled. To solve this question we can use equation \eqref{ratios} to solve for \(v'\), explicitly

\begin{equation}
\frac{v'}{v}=\sqrt{\frac{T'}{T}},
\end{equation}

\begin{equation}
v'=v\sqrt{\frac{T'}{T}}.
\end{equation}

The new tension is three times the original tension, thus \(T'=3T\) and the equation above is simplified to

\begin{equation}
v'=v\sqrt{\frac{3T}{T}},
\end{equation}

\begin{equation}
v'=v\sqrt{3}.
\end{equation}

Using the numerical values given by the problem we get

\begin{equation}
v'=(50\,\text{m/s})\sqrt{3},
\end{equation}

\begin{equation}
v'\approx 86.6\,\text{m/s}.
\end{equation}