### Video Transcript

A wave is modeled with the function ๐ฆ, a function of ๐ฅ and ๐ก, equals 0.25 meters times the cosine of 0.30๐ฅ minus 0.90๐ก plus ๐ divided by three, where ๐ฅ is measured in meters and ๐ก is measured in seconds. What is the waveโs amplitude? What is the waveโs wave number? What is the waveโs angular frequency? What is the waveโs speed? What is the phase shift of the wave? What is the waveโs wavelength? What is the waveโs period?

In this exercise, weโll work with the function ๐ฆ, a function of position and time, to answer these questions. Weโll call the amplitude of the wave ๐ด. Weโll call the wave number ๐. The waveโs angular frequency weโll call ๐. Wave speed weโll refer to as ๐ฃ. Phase shift weโll call ๐. The wavelength weโll call ๐. And the waveโs period weโll call capital ๐.

Letโs look more closely at this function in order to answer these questions. This wave equation has a particular form which is familiar for waves on a string. We can recall the general form of that equation for the waveโs position as a function of time. Wave position, ๐ฅ, as a function of time, ๐ก, is modeled in general by the equation which says ๐ฆ equals the amplitude ๐ด times the sine of the wavenumber ๐ times the position ๐ฅ minus the angular frequency ๐ times time ๐ก plus ๐, the phase angle of the wave.

Studying this general form of wave position in time helps us solve for the first three variables the problem statement asks us to find. 0.25 meters must be the wave amplitude ๐ด. Then moving along to the right in the equation, 0.30 inverse meters must be the waveโs wave number ๐. Continuing onto the right, 0.90 must be ๐ in units of inverse seconds. And continuing on, we see that ๐ divided by three is this waveโs phase shift or phase angle, ๐.

Weโve been able to solve for much about the waveโs properties by studying the general equation for a transverse wave. Now letโs recall several more wave relationships. We know that the speed of a wave, ๐ฃ, is equal to the wave frequency times its wavelength. And that frequency, ๐, is equal to angular frequency, ๐, divided by two ๐. And also that wavelength, ๐, is equal to two ๐ divided by the wavenumber, ๐.

For our purposes, this means that wave speed, ๐ฃ, is equal to ๐ divided by two ๐ times two ๐ divided by ๐. Or with the two ๐ factors canceling out, ๐ divided by ๐. When we enter in the values for these two variables and calculate this fraction, we find that the wave speed is 3.0 meters per second.

Having solved for ๐ฃ, weโve encountered a relationship that can help us solve for ๐. The wavelength of the wave is equal to two ๐ divided by the wave number ๐. Plugging in the value for ๐ we solved for earlier, when we calculate this fraction, we find that, to two significant figures, the wavelength ๐ is 21 meters. Finally, we move on to the period of the wave ๐.

To solve for the period, letโs recall the relationship between ๐ and frequency, ๐. The wave period, ๐, is equal to the inverse of ๐, one over the frequency. So in our situation, ๐ equals two ๐ over the angular frequency, ๐, which is two ๐ divided by 0.90 inverse seconds. To two significant figures, this equals 7.0 seconds. Thatโs the period of this wave, and these are all the wave properties derived from the wave equation and from general wave relationships.