Video: Analyzing the Wave Function of a Traveling Wave

A transverse wave on a horizontal string with linear mass density of 0.00600 kg/m is described with the equation 𝑦(π‘₯, 𝑑) = 0.300 m sin((2πœ‹/4.00)(π‘₯ βˆ’ 𝑣_(πœ”) 𝑑)), where 𝑣_πœ” is the wave’s angular frequency, π‘₯ is measured in meters, and 𝑑 is measured in seconds. The string is under a tension of 3.00 Γ— 10Β² N. At what speed does the wave propagate along the string? What is the wave number of the wave? What is the angular frequency of the wave?

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Video Transcript

A transverse wave on a horizontal string with linear mass density of 0.00600 kilograms per meter is described with the equation 𝑦 as a function of π‘₯ and 𝑑 equals 0.300 meters times the sin of two πœ‹ divided by 4.00 times the quantity π‘₯ minus 𝑣 sub πœ” times 𝑑, where 𝑣 sub πœ” is the wave’s angular frequency, π‘₯ is measured in meters and 𝑑 is measured in seconds. The string is under a tension of 3.00 times 10 to the two newtons. At what speed does the wave propagate along the string? What is the wave number of the wave? What is the angular frequency of the wave?

Let’s start by highlighting some of the important information we’re told about the scenario. We’re told that the string has a linear mass density of 0.00600 kilograms per meter. And that the string is described using the equation 𝑦 as a function of π‘₯ and 𝑑. We’re also told that the string is under a tension of 3.00 times 10 to the two newtons. We want to know the speed at which the wave propagates along the string, which we’ll call 𝑣. We want to know the wave number of the wave, which we’ll call π‘˜. And we want to know the angular frequency of the wave, which is 𝑣 sub πœ”.

The first variable we want to solve for is the wave speed 𝑣. And to do that, we’ll rely on a relationship between wave speed when a wave is on a string, the tension on the string, and the string’s mass per unit length which we’ve called πœ‡. Wave speed is given by the square root of 𝐹 sub 𝑑, the tension force, divided by πœ‡. When we apply this relationship to our scenario, we see we’ve been given 𝐹 sub 𝑑, the tension force, and πœ‡. When we plug-in those values and calculate that square root, we find that the wave speed 𝑣 is equal to 224 meters per second. That’s how fast the waves travel along the string.

Now we move on to solving for π‘˜, the wave number. To do this, we’ll recall the standard mathematical form of a transverse wave. This standard form, given as a function of position π‘₯ and time 𝑑, shows us that if we compare this form with our equation, 𝑦 as a function of π‘₯ and 𝑑, then we can see that 0.300 meters is the amplitude 𝐴 of the wave and that two πœ‹ divided by 4.00 is the wave number π‘˜. So π‘˜ equals two πœ‹ over 4.00 inverse meters or 1.57 inverse meters. This is the wave number of the wave.

Finally, we want to solve for the angular frequency of the wave 𝑣 sub πœ”. To do that, we can recall three relationships between wave variables. First, we recall that the angular frequency πœ” is equal to the linear frequency 𝑓 times two πœ‹. Next, we can recall that the speed of a wave 𝑣 is equal to its frequency times its wavelength. Which means that frequency equals wave speed 𝑣 divided by wavelength πœ†. And finally, we recall that πœ† is equal to two πœ‹ divided the wave number π‘˜. And when we substitute that in for πœ† in our equation, we see that the two πœ‹ terms cancel out. And we’re left with πœ”, or angular frequency, equals wave speed 𝑣 times wave number π‘˜. Since we solved for these values in parts one and two, we can plug them in now. When we multiply these two terms together, we find that angular frequency is 142 inverse seconds. That’s the angular frequency of the wave travelling along the string.

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