A wave is modeled with the function 𝑦, a function of 𝑥 and 𝑡, equals 0.25 meters times the cosine of 0.30𝑥 minus 0.90𝑡 plus 𝜋 divided by three, where 𝑥 is measured in meters and 𝑡 is measured in seconds. What is the wave’s amplitude? What is the wave’s wave number? What is the wave’s angular frequency? What is the wave’s speed? What is the phase shift of the wave? What is the wave’s wavelength? What is the wave’s period?
In this exercise, we’ll work with the function 𝑦, a function of position and time, to answer these questions. We’ll call the amplitude of the wave 𝐴. We’ll call the wave number 𝑘. The wave’s angular frequency we’ll call 𝜔. Wave speed we’ll refer to as 𝑣. Phase shift we’ll call 𝜙. The wavelength we’ll call 𝜆. And the wave’s period we’ll call capital 𝑇.
Let’s 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 wave’s position as a function of time. Wave position, 𝑥, as a function of time, 𝑡, is modeled in general by the equation which says 𝑦 equals the amplitude 𝐴 times the sine of the wavenumber 𝑘 times the position 𝑥 minus the angular frequency 𝜔 times time 𝑡 plus 𝜙, 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 𝐴. Then moving along to the right in the equation, 0.30 inverse meters must be the wave’s wave number 𝑘. Continuing onto the right, 0.90 must be 𝜔 in units of inverse seconds. And continuing on, we see that 𝜋 divided by three is this wave’s phase shift or phase angle, 𝜙.
We’ve been able to solve for much about the wave’s properties by studying the general equation for a transverse wave. Now let’s recall several more wave relationships. We know that the speed of a wave, 𝑣, is equal to the wave frequency times its wavelength. And that frequency, 𝑓, is equal to angular frequency, 𝜔, divided by two 𝜋. And also that wavelength, 𝜆, is equal to two 𝜋 divided by the wavenumber, 𝑘.
For our purposes, this means that wave speed, 𝑣, is equal to 𝜔 divided by two 𝜋 times two 𝜋 divided by 𝑘. Or with the two 𝜋 factors canceling out, 𝜔 divided by 𝑘. 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 𝑣, we’ve encountered a relationship that can help us solve for 𝜆. The wavelength of the wave is equal to two 𝜋 divided by the wave number 𝑘. Plugging in the value for 𝑘 we solved for earlier, when we calculate this fraction, we find that, to two significant figures, the wavelength 𝜆 is 21 meters. Finally, we move on to the period of the wave 𝑇.
To solve for the period, let’s recall the relationship between 𝑇 and frequency, 𝑓. The wave period, 𝑇, is equal to the inverse of 𝑓, one over the frequency. So in our situation, 𝑇 equals two 𝜋 over the angular frequency, 𝜔, which is two 𝜋 divided by 0.90 inverse seconds. To two significant figures, this equals 7.0 seconds. That’s the period of this wave, and these are all the wave properties derived from the wave equation and from general wave relationships.