Video: Determining the Properties of a Sound Wave Using a Wave Function

Consider a sound wave modeled with the equation 𝑠(π‘₯, 𝑑) = 4.00 nm cos (3.66π‘₯ βˆ’ 1256𝑑), where π‘₯ is measured in meters and 𝑑 is measured in seconds. What is the maximum displacement of the sound wave? What is the wavelength? What is the frequency of the sound wave? What is the speed of the sound wave?

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Video Transcript

Consider a sound wave modeled with the equation 𝑠 as a function of π‘₯ and 𝑑 equals 4.00 nanometers times the cosine of 3.66 times π‘₯ minus 1256 times 𝑑, where π‘₯ is measured in meters and 𝑑 is measured in seconds. What is the maximum displacement of the sound wave? What is the wavelength? What is the frequency of the sound wave? What is the speed of the sound wave?

In this multipart problem, we’ll call that maximum displacement of the sound wave 𝐴. We’ll call the wavelength πœ†, the frequency 𝑓, and the speed of the sound wave 𝑣. To begin, let’s recall the general wave equation, which is a function of both position π‘₯ and time 𝑑. In general, the amplitude of a wave, 𝑦, as a function of the wave position and time, 𝑑, is equal to its maximum amplitude, capital 𝐴, times the sin of the wave number π‘˜ times its position π‘₯ minus the angular frequency πœ” times time 𝑑.

If we apply this wave equation to the equation we’ve been given, 𝑠 as a function of π‘₯ and 𝑑, we see that 4.00 nanometers is in the place of amplitude, 𝐴; 3.66 is in the place of wavenumber π‘˜; and 1256 occupies in the place of πœ”, the angular frequency. For our wave then, the amplitude 𝐴 is 4.00 nanometers, which is the answer to the first part of the problem.

When we move on to part two to solve for the wavelength πœ†, we can recall that there is a mathematical relationship between wavelength πœ† and wavenumber π‘˜: πœ† equals two πœ‹ divided by π‘˜. In our case, π‘˜ is equal to 3.66 inverse meters. And when we calculate this fraction, we find that πœ† is equal to 1.72 meters. That’s the wavelength of the wave derived from the wave equation.

As we move on to solving for the frequency 𝑓, let’s recall the relationship between linear frequency 𝑓 and angular frequency πœ”: 𝑓 is equal to πœ” divided by two πœ‹. In our scenario, πœ” is equal to 1256 inverse seconds, and this fraction gives us a frequency of 200 hertz. That’s the frequency of the waves given by this wave equation.

We’re now ready to move on to the last part of the problem, solving for wave speed 𝑣. To do that, let’s recall a relationship between wave speed, wave frequency, and wavelength. The speed of a wave 𝑣 is equal to frequency times wavelength. Since we’ve solved for frequency and wavelength in previous parts, we can insert those values here. To three significant figures, the speed of this wave is 343 meters per second, a standard assumed value for the speed of sound in air.

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