Video: Finding the Wavelength of Mechanical Waves in Superposition

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string, producing a standing wave. The linear mass density of the string is 0.075 kg/m and the tension in the string is 5.0 N. The time interval between instances of total destructive interference is 0.13 s. What is the wavelength of the waves?

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Video Transcript

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string, producing a standing wave. The linear mass density of the string is 0.075 kilograms per meter and the tension in the string is 5.0 newtons. The time interval between instances of total destructive interference is 0.13 seconds. What is the wavelength of the waves?

In this problem, we’ll assume that no energy is lost to moving the string, that the waves continue to propagate along it with no loss of amplitude.

We’re told in this problem statement that the linear mass density of the string is equal to 0.075 kilograms per meter. We’ll call that the Greek letter 𝜇.

We’re also told that the tension in the string is 5.0 newtons. We’ll refer to that tension as capital 𝑇. And finally, we’re told that there are 0.13 seconds in between instances of total destructive interference along the string. We want to know what is the wavelength, 𝜆, of the waves.

Let’s start by drawing a diagram of these two waves along the string. Here we have a diagram of our two waves that interact with one another along a common string. Because of the interference of waves, these two waves traveling in opposite directions create a standing wave.

At the particular moment in time at which we’ve drawn the waves, they would interfere perfectly destructively and create a standing wave with an amplitude of zero.

We’re told that this happens every 0.13 seconds. We’ll call that time 𝑡 sub 𝑑. That’s how much time it takes each wave to move one-half wavelength and recreate the perfectly destroyed pattern.

To start solving for wavelength, let’s recall a relationship for wave speed on a string. The speed of a wave, 𝑣, is equal to the square root of the tension of the string divided by its mass per unit length, 𝜇.

When it comes to the speed of a wave, we also know that that wave speed is equal to the wavelength of the wave multiplied by its frequency.

If we combine these two equations for wave speed in our scenario, we see that the square root of the string tension divided by its mass per unit length is equal to the wavelength of the wave times its frequency.

If we multiply both sides by one over 𝑓, then the frequency term on the right side of the equation cancels out, leaving us with an equation for 𝜆, the wavelength.

In the problem statement, we’re given 𝑇 and 𝜇, but what about 𝑓?

Let’s look back at our drawing. We’re told that, in 0.13 seconds, each one of the waves moves a distance, so that when it combines with the other wave, destructive interference occurs.

We can see that, for each wave, this involves moving one-half wavelength forward in time. Wave frequency is defined as the number of wavelengths passing a point each second. So if one-half of a wavelength passes a point every 0.13 seconds, that means that an entire wavelength will pass a point every 0.26 seconds; that is, we’ve multiplied both sides by two.

This tells us that the frequency of the waves on the string is equal to one divided by 0.26 seconds. Knowing that, we now have all the information we need to solve for the wavelength, 𝜆.

Let’s enter those values into our equation for 𝜆 now. When we enter these values into our calculator, to two significant figures the wavelength, 𝜆, is equal to 2.1 meters. That’s the wavelength of each wave on the string.

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