Video: Determining the Frequency of Mechanical Waves on a String

A string has a linear mass density πœ‡ = 0.00700 kg/m, a length 𝐿 = 0.700 m, a tension of 𝐹_𝑇 = 110 N, and oscillates in a mode 𝑛 = 3. What is the frequency of the oscillations?

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Video Transcript

A string has a linear mass density πœ‡ equals 0.00700 kilograms per meter, a length 𝐿 equals 0.700 meters, a tension of 𝐹 sub 𝑇 equals 110 newtons and oscillates in a mode 𝑛 equals three. What is the frequency of the oscillations?

We’re told in this problem statement that πœ‡ is 0.00700 kilograms per meter, that the string has a length of 0.700 meters, that it’s under a tension of 110 newtons, and that it oscillates in the mode 𝑛 equals three. We want to solve for the oscillation frequency, which we’ll call 𝑓.

To start off, let’s draw a diagram of the scenario. In this situation, we have a string fixed at both ends. The string is under tension 𝐹 sub 𝑇, and it oscillates in the 𝑛 equals three mode, which means that from one end to the other they are one and a half wavelengths on the string.

To begin solving for the frequency, let’s recall two equations for the speed of a wave. First when a wave is on a string, the wave speed 𝑉 is equal to the square root of the tension that the string is under, 𝐹 sub 𝑇, divided by the string’s mass per unit length, πœ‡. And, second we recall that in general for a wave, the wave speed 𝑉 is equal to the wave frequency 𝑓 times wavelength πœ†.

When we apply these two relationships to our situation, we see they can be combined to write that the square root of 𝐹 sub 𝑇 over πœ‡ equals 𝑓 times πœ† or 𝑓 equals one over πœ† times the square root of 𝐹 sub 𝑇 over πœ‡. If we look again in our diagram, we can see that there’s a relationship between the length of the string 𝐿 and the wavelength πœ†.

Because one and a half wavelengths fit along the string at any given time in this mode, 𝐿 is equal to three halves πœ† or πœ† equals two 𝐿 divided by three. We can substitute this expression for πœ† into our equation for frequency.

Since we’ve been given 𝐿, πœ‡, and 𝐹 sub 𝑇 in the problem statement, we can plug in for those values now. When we enter these values into a calculator, we find that, to three significant figures, the frequency of oscillation is 269 hertz. That’s how many wavelengths of this wave pass a given point on the string every second.

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