Mass on a Rope

a) Write the expression for a propagating wave, and compare the coefficients for the wavelength, the frequency, and the velocity. Then, with the length, the time can be found.

b) Based on Newton's Second Law, find the tension in terms of the mass \(M\). Then, with the equation for the velocity of a wave traveling through a medium, substitute the tension and solve for \(M\).

a) The equation for a travelling wave is:

\begin{equation*}
y(x,t)=A\sin\left(\frac{2\pi}{\lambda}x+2\pi f t\right),
\end{equation*}

where the coefficient of the \(x\) variable can lead us to the wavelength:

\begin{equation*}
100\,\text{m}^{-1}=\frac{2\pi}{\lambda},
\end{equation*}

and the coefficient of the \(t\) variable will relate us the frequency:

\begin{equation*}
3000\,\text{s}^{-1}=2\pi f.
\end{equation*}

Since, the velocity is \(v = \lambda f\), and the time can be found as \(t' = \frac{L}{v}\). Finding the velocity with the wavelength and the frequency, and using the numerical values for the length we get:

\begin{equation*}
t' \approx 0.067' \, \text{s}.
\end{equation*}

b) Based on Newton's Second Law, we have:

\begin{equation*}
T-Mg = 0.
\end{equation*}

Since the velocity is \(v = \sqrt{\frac{TL}{m}} \), then substituting the tension and solving for the mass \(M\), we obtain:

\begin{equation*}
M=\frac{mv^2}{gL}.
\end{equation*}

Plugging in the numerical values, we get:

\begin{equation*}
M\approx 23\,\text{kg}.
\end{equation*}

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

a) Let us calculate the time it takes for a pulse to go through the entire string. For this exercise we will assume that the mass \(M\) is much larger than the mass of the rope \(m\) and so the tension along the rope is constant. We will first find the wavelength and frequency of the wave and then its velocity. With the velocity and length of the rope we will be able to find the time it takes for a pulse to travel the whole rope. A sinusoidal wave travelling along the negative X axis and oscillating along the Y axis can be described as a function of position \(x\) and time \(t\) using the following expression

\begin{equation}
\label{wave}
y(x,t)=A\sin\left(\frac{2\pi}{\lambda}x+2\pi f t\right),
\end{equation}

where \(A\) is the amplitude of the wave, \(\lambda\) is the wavelength of the wave and \(f\) its frequency. Contrasting equation \eqref{Eq:1} with \eqref{wave}, in particular the term that multiplies \(x\) we deduce that

\begin{equation}
100\,\text{m}^{-1}=\frac{2\pi}{\lambda},
\end{equation}

which solving for \(\lambda\) is

\begin{equation}
\label{lambda}
\lambda=\frac{2\pi}{100\,\text{m}^{-1}}.
\end{equation}

We can also deduce by contrasting equations \eqref{Eq:1} and \eqref{wave}, focusing on the term that multiplies time \(t\) that

\begin{equation}
3000\,\text{s}^{-1}=2\pi f.
\end{equation}

Solving for the frequency we get

\begin{equation}
\label{frec}
f=\frac{3000\,\text{s}^{-1}}{2\pi}.
\end{equation}
Now that we have explicit expressions for the wavelength and frequency we can calculate the velocity of the wave \(v\) using the relation

\begin{equation}
v=\lambda f,
\end{equation}

and using expressions in equations \eqref{lambda} and \eqref{frec}

\begin{equation}
v=\left(\frac{2\pi}{100\,\text{m}^{-1}}\right)\left(\frac{3000\,\text{s}^{-1}}{2\pi}\right),
\end{equation}

\begin{equation}
v=30\,\text{m/s}.
\end{equation}

If the rope measures \(L\), then the time \(t'\) it takes for the pulse to travel from one end to the other of the rope is given by the kinematic expression

\begin{equation}
t'=\frac{L}{v},
\end{equation}

which after using the numerical values is

\begin{equation}
t'=\frac{2\,\text{m}}{30\,\text{m/s}},
\end{equation}

\begin{equation}
t'\approx 0.067' \,\text{s}.
\end{equation}

b) For the final part of this problem we need to find the mass \(M\). For a string of mass \(m\), length \(L\) and tension \(T\) we know that the velocity at which a wave propagates through it is

\begin{equation}
\label{velo}
v=\sqrt{\frac{TL}{m}}.
\end{equation}

The mass \(m\) and length \(L\) are known o we must find an expression for the tension \(T\). If we make a freebody diagram for the mass \(M\), which we consider to be at equilibrium we only have two forces: the weight \(Mg\) and the tension \(T\).

MAKE FREEBODY DIAGRAM OF THE BOX HANGED BY THE ROPE. DRAW THE TENSION AND WEIGHT VECTORS. REMEMBER THE COORDINATE SYSTEM IS SUCH THAT X POINTS DOWNWARDS
Using Newton's second law for the static case along the vertical axis we obtain

\begin{equation}
Mg\,\hat{\textbf{i}}-T\,\hat{\textbf{i}}=0,
\end{equation}

where we can drop the vector notation and write

\begin{equation}
Mg-T=0.
\end{equation}

Solving for the tension we have

\begin{equation}
T=Mg,
\end{equation}

where \(g\) is the graviational acceleration of Earth. Using this expression for tension in equation \eqref{velo} we obtain

\begin{equation}
v=\sqrt{\frac{MgL}{m}},
\end{equation}

where now the only unknown is \(M\). Elevating both sides to the power of two we get

\begin{equation}
v^2=\frac{MgL}{m},
\end{equation}

and solving for \(M\) we have

\begin{equation}
M=\frac{mv^2}{gL}.
\end{equation}

Using the numerical value we get

\begin{equation}
M=\frac{(0.5\,\text{kg})(30\,\text{m/s})^2}{(9.8\,\text{m/s}^2)(2\,\text{m})},
\end{equation}

\begin{equation}
M\approx 23\,\text{kg},
\end{equation}

which is much larger than \(m=0.5\,\text{kg}\) as we assume from the beginning.