# A traveling wave along the x-axis is given by the following wave function.

Here, $x$ and $\Psi$ are measured in meters while $t$ is in seconds. Carefully study this wave equation and calculate the following quantities:

$\boldsymbol{ \Psi(x,t) = 4.8 cos ( 1.2x – 8.2t + 0.54 ) }$

– Frequency ( in hertz )

– Wavelength ( in meters )

– Wave speed ( in meters per second )

– Phase angle ( in radians )

The aim of this question is to develop an understanding of the traveling wave equation.

To solve this question, we simply compare the given equation with the standard wave equation and then find the necessary parameters as given below:

$\Psi(x,t) = A cos ( k x – \omega t + \phi )$

Then we simply find wavelength, speed and frequency by following these formulas:

$f = \frac{ \omega }{ 2 \pi }$

$\lambda = \frac{ 2 \pi }{ k }$

$v = f \cdot \lambda$

Step 1: Given the function:

$\Psi(x,t) = 4.8 \ cos ( 1.2x \ – \ 8.2t \ + \ 0.54 )$

The standard wave equation is given by:

$\Psi(x,t) = A \ cos ( k x \ – \ \omega t \ + \ \phi )$

Comparing the give equation with the standard equation, we can see that:

$A = 4.8$

$k = 1.2$

$\omega = 8.2 \ \frac{rad}{sec}$

$\phi = 0.54 \ rad$

Step 2: Calculating Frequency:

$f = \frac{ \omega }{ 2 \pi }$

$f = \dfrac{ 8.2 \ \frac{rad}{sec} }{ 2 \pi \ rad}$

$f = 0.023 \ sec^{-1}$

Step 3: Calculating Wavelength:

$\lambda = \frac{ 2 \pi }{ k }$

$\lambda = \frac{ 2 \pi }{ 1.2 }$

$\lambda = 300 \ meter$

Step 4: Calculating Wave Speed:

$v = f \cdot \lambda$

$v = ( 0.023 \ sec^{-1}) ( 300 \ meter )$

$v = 6.9 \ \frac{meter}{sec}$

## Numerical Result

For the given wave equation:

– Frequency ( in hertz ) $\boldsymbol{ f = 0.023 \ sec^{-1} }$

– Wavelength ( in meters ) $\boldsymbol{ \lambda = 300 \ meter }$

– Wave speed ( in meters per second ) $\boldsymbol{ v = 6.9 \ \frac{meter}{sec} }$

– Phase angle ( in radians ) $\boldsymbol{ \phi = 0.54 \ rad }$

## Example

Find Frequency (in hertz), Wavelength (in meters), Wave speed (in meters per second) and Phase angle (in radians) for the following wave equation:

$\Psi(x,t) = 10 cos ( x – t + \pi )$

Comparing with the standard equation, we can see that:

$A = 10 , \ k = 1, \ \omega = 1 \frac{rad}{sec}, \ \phi = \pi \ rad$

Calculating Frequency:

$f = \frac{ \omega }{ 2 \pi } = \dfrac{ 1 \ \frac{rad}{sec} }{ 2 \pi \ rad} = \frac{1}{ 2 \pi } \ sec^{-1}$

Calculating Wavelength:

$\lambda = \frac{ 2 \pi }{ k } = \frac{ 2 \pi }{ 1 } = 2 \pi \ meter$

Calculating Wave Speed:

$v = f \cdot \lambda = ( \frac{1}{ 2 \pi } sec^{-1}) ( 2 \pi meter ) = 1 \ \frac{m}{s}$