Question Video: Determining the Wave Parameters from a Wave Function

A wave is modeled by the wavefunction 𝑦(π‘₯, 𝑑) = (0.30 m) sin [(2πœ‹)/(4.50) (π‘₯ βˆ’ 18.00𝑑)], where π‘₯ is measured in meters and 𝑑 is measured in seconds. Find the amplitude of the wave. Find the wavelength of the wave. Find the speed of propagation of the wave. Find the frequency of the wave. Find the period of the wave.

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Video Transcript

A wave is modeled by the wavefunction 𝑦 as a function of π‘₯ and 𝑑 is equal to 0.30 meters times the sin of two πœ‹ over 4.50 times the quantity π‘₯ minus 18.00𝑑, where π‘₯ is measured in meters and 𝑑 is measured in seconds. Find the amplitude of the wave. Find the wavelength of the wave. Find the speed of propagation of the wave. Find the frequency of the wave. Find the period of the wave.

We can label our wave amplitude capital 𝐴, the wavelength πœ†, the wave speed 𝑣, the wave frequency 𝑓, and the period capital 𝑇. We can start solving for these various wave properties by recalling that a one-dimensional transverse wave is modeled by the function 𝑦 is a function of π‘₯ and 𝑑 is equal to wave amplitude times the sine of wavenumber π‘˜ times position minus angular frequency of the wave times time.

Looking back at our given wave function, we see right away that the amplitude of our wave is given as 0.30 meters. To solve next for the wavelength of our wave πœ†, we can recall that wavelength and wavenumber are related. Wavenumber π‘˜ is equal to two πœ‹ over wavelength. And as we look at our given function, we see that this is two πœ‹ over 4.50. This indicates that our wavelength πœ† is 4.50 meters.

Next, we want to solve for wave speed 𝑣. And to do this, we’ll recall two pieces of information. First, we recall that angular frequency is equal to two πœ‹ times frequency 𝑓. Working off of our given wave function, we see that two πœ‹ times 18.00 divided by 4.50 is equal to πœ”, which is two πœ‹ times 𝑓. The factors of two πœ‹ cancel out. And 𝑓 is equal to 4.00 hertz. We can fill that in for our frequency 𝑓 in advance. But now we wanna use the relationship that says 𝑣 is equal to πœ† times 𝑓 to solve for 𝑣. 𝑣 is equal to πœ†π‘“ or 4.50 meters times 4.00 hertz, which is equal to 18.00 meters per second. That’s the speed of propagation of this wave.

Finally, we want to solve for the wave period 𝑇. Recalling that wave period is equal to the inverse of wave frequency, the period of this wave is equal to one over 4.00 hertz or 0.250 seconds. That’s the wave period determined using the wave function for a wave as well as relationships among wave parameters.

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