Wave Interference

a) Compare the coefficients for a general function sinusoidal wave propagating to find each parameter.

b) Use a trigonometric identity to get the combined function.

a) The transverse sinusoidal wave propagating can be written as:

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

The wavelengths are obtained from the coefficient of the \(x\) variable. Then:

\begin{equation*}
\lambda_1 \approx 1.26 \,\text{m},
\end{equation*}

and

\begin{equation*}
\lambda_2 \approx 1.57 \,\text{m}.
\end{equation*}

The same analysis can be made for the factors that multiply the time \(t\). Then:

\begin{equation*}
T_1 \approx 1.26\,\text{s},
\end{equation*}

and

\begin{equation*}
T_2 \approx 0.63 \,\text{s}.
\end{equation*}

The frequency \(f\) is \(f = 1/T\). Then:

\begin{equation*}
f_1 \approx  1.59 \,\text{Hz},
\end{equation*}

and

\begin{equation*}
f_2\approx  0.79\,\text{Hz}.
\end{equation*}

The velocity \(v= \lambda f\). Then:

\begin{equation*}
v_1=2.00 \,\text{m/s},
\end{equation*}

and

\begin{equation*}
v_2=1.25\,\text{m/s},
\end{equation*}

b) The trigonometric identity useful here is:

\begin{equation*}
\sin(C)+\sin(D)=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right).
\end{equation*}

Then, the result is:

\begin{equation*}
y_{\text{total}}(x,t)=3.0\sin\left(\frac{9}{2}x-\frac{15}{2}t\right)\cos\left(\frac{x-5t}{2}\right).
\end{equation*}

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

a) For the first part, we need to find different physical properties such as amplitude, period and velocity for each of the waves. In general, for a transverse sinusoidal wave propagating in the positive X direction and oscillating in the Y axis we can write

\begin{equation}
\label{canon}
y(x,t)=A\sin\left(\frac{2\pi}{\lambda}x-\frac{2\pi}{T}t+\delta\right),
\end{equation}

where \(A\) is the amplitude, \(\lambda\) is the wavelength and \(T\) is the period and \(\delta\) is a phase angle that depends on the initial conditions for the wave. Let us now write both \(y_1(x,t)\) and \(y_2(x,t)\) in the general form as in equation \eqref{canon}. Let us start by \(y_1(x,t)\), first let us use the trigonometric property of the sine function \(-\sin(x)=\sin(-x)\) to write

\begin{equation}
y_1(x,t)=1.5\sin\left(-10(t-x/2)\right),
\end{equation}

which after some algebra and expanding the parenthesis becomes

\begin{equation}
\label{y1canon}
y_{1}(x,t)=1.5\sin\left(5x-10t\right).
\end{equation}

Contrasting equation \eqref{canon} with equation \eqref{y1canon} and using the sub-index 1 in all the variables corresponding to \(y_1(x,t)\) we notice that \(A_1=1.5\,\text{m}\). Equating the factors that multiply the \(x\) we get

\begin{equation}
\frac{2\pi}{\lambda_1}=5,
\end{equation}

from which we can solve for \(\lambda_1\) to obtain

\begin{equation}
\lambda_1=\frac{2\pi}{5}\approx 1.26\,\text{m}.
\end{equation}

The same analysis can be made for the factors that multiply the time \(t\), then

\begin{equation}
\frac{2\pi}{T_1}=10,
\end{equation}

which solving for the period \(T_1\) we have

\begin{equation}
T_1=\frac{2\pi}{10}\approx 0.63\,\text{s}.
\end{equation}

The frequency \(f\) can be obtained through the period \(T\) using the relation

\begin{equation}
f=\frac{1}{T}.
\end{equation}

Thus the frequency of the wave described by \(y_1(x,t)\) is

\begin{equation}
f_1=\frac{1}{T_1},
\end{equation}

\begin{equation}
f_1\approx \frac{1}{0.63\,\text{s}}=1.59\,\text{Hz}.
\end{equation}

The velocity \(v\) of propagation of the sinusoidal wave can be calculated in terms of the wavelength and period as

\begin{equation}
v=\frac{\lambda}{T}.
\end{equation}

Then for the wave described by \(y_1(x,t)\) we get

\begin{equation}
v_1=\frac{\lambda_1}{T_1},
\end{equation}

\begin{equation}
v_1=\frac{1.26\,\text{m}}{0.63\,\text{s}}=2.00\,\text{m/s}.
\end{equation}

We can continue with the wave described by \(y_2(x,t)\) now. Notice that if we write \(y_2(x,t)\)

\begin{equation}
\label{y2canon}
y_2(x,t)=1.5\sin\left(4x-5t\right),
\end{equation}

it is already in the form of equation \eqref{canon}. Contrasting equation \eqref{canon} with equation \eqref{y2canon} and using the sub-index 2 in all the variables corresponding to \(y_2(x,t)\) we notice that \(A_2=1.5\,\text{m}\). Equating the factors that multiply the \(x\) we get

\begin{equation}
\frac{2\pi}{\lambda_2}=4,
\end{equation}

from which we can solve for \(\lambda_2\) to obtain

\begin{equation}
\lambda_2=\frac{2\pi}{4}\approx 1.57\,\text{m}.
\end{equation}

The same analysis can be made for the factors that multiply the time \(t\), then

\begin{equation}
\frac{2\pi}{T_2}=5,
\end{equation}

which solving for the period \(T_2\) we have

\begin{equation}
T_2=\frac{2\pi}{10}\approx 1.26\,\text{s}.
\end{equation}

The frequency \(f\) can be obtained through the period \(T\) using the relation

\begin{equation}
f=\frac{1}{T}.
\end{equation}

Thus the frequency of the wave described by \(y_2(x,t)\) is

\begin{equation}
f_2=\frac{1}{T_2},
\end{equation}

\begin{equation}
f_2\approx \frac{1}{1.26\,\text{s}}=0.79\,\text{Hz}.
\end{equation}

The velocity \(v\) of propagation of the sinusoidal wave can be calculated in terms of the wavelength and period as

\begin{equation}
v=\frac{\lambda}{T}.
\end{equation}

Then for the wave described by \(y_1(x,t)\) we get

\begin{equation}
v_2=\frac{\lambda_2}{T_2},
\end{equation}

\begin{equation}
v_1=\frac{1.57\,\text{m}}{1.26\,\text{s}}=1.25\,\text{m/s}.
\end{equation}

Because of the minus sign in the term that multiplies the \(t\) in both waves, the direction of propagation is to the right along the X positive axis.

b) Using the superposition principle, the combined disturbance will be described by the sum of the functions describing both waves, namely

\begin{equation}
y_{\text{total}}(x,t)=y_1(x,t)+y_2(x,t).
\end{equation}

Using the explicit expressions for \(y_1(x,t)\) and \(y_2(x,t)\) given in \eqref{y1canon} and \eqref{y2canon} we obtain

\begin{equation}
y_{\text{total}}(x,t)=1.5\sin\left(5x-10t\right)+1.5\sin\left(4x-5t\right),
\end{equation}

where the term \(1.5\) can be factorized to get

\begin{equation}
\label{ytot}
y_{\text{total}}(x,t)=1.5\left(\sin\left(5x-10t\right)+\sin\left(4x-5t\right)\right).
\end{equation}

We will now use the trigonometric identity

\begin{equation}
\sin(C)+\sin(D)=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right),
\end{equation}

into equation \eqref{ytot} to write

\begin{equation}
\label{ytot2}
y_{\text{total}}(x,t)=1.5\left(2\sin\left(\frac{5x-10t+4x-5t}{2}\right)\cos\left(\frac{5x-10t-(4x-5t)}{2}\right)\right).
\end{equation}

After some simplification equation \eqref{ytot2} becomes

\begin{equation}
\label{ytot3}
y_{\text{total}}(x,t)=3.0\sin\left(\frac{9}{2}x-\frac{15}{2}t\right)\cos\left(\frac{x-5t}{2}\right).
\end{equation}

The form of the waves is presented in the following figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/sumondas.png}
\caption{From top to bottom \(y_1\) ,\(y_2\) and \(y_{\text{total}}\) at \(t=0\)}.
\label{fig:sumondas}
\end{figure}