# Video: Determining the Frequency of Mechanical Waves on a String

A string has a linear mass density π = 0.00700 kg/m, a length πΏ = 0.700 m, a tension of πΉ_π = 110 N, and oscillates in a mode π = 3. What is the frequency of the oscillations?

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### Video Transcript

A string has a linear mass density π equals 0.00700 kilograms per meter, a length πΏ equals 0.700 meters, a tension of πΉ sub π equals 110 newtons and oscillates in a mode π equals three. What is the frequency of the oscillations?

Weβre told in this problem statement that π is 0.00700 kilograms per meter, that the string has a length of 0.700 meters, that itβs under a tension of 110 newtons, and that it oscillates in the mode π equals three. We want to solve for the oscillation frequency, which weβll call π.

To start off, letβs draw a diagram of the scenario. In this situation, we have a string fixed at both ends. The string is under tension πΉ sub π, and it oscillates in the π 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, letβs recall two equations for the speed of a wave. First when a wave is on a string, the wave speed π is equal to the square root of the tension that the string is under, πΉ sub π, divided by the stringβs mass per unit length, π. And, second we recall that in general for a wave, the wave speed π is equal to the wave frequency π times wavelength π.

When we apply these two relationships to our situation, we see they can be combined to write that the square root of πΉ sub π over π equals π times π or π equals one over π times the square root of πΉ sub π over π. If we look again in our diagram, we can see that thereβs a relationship between the length of the string πΏ and the wavelength π.

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

Since weβve been given πΏ, π, and πΉ sub π 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. Thatβs how many wavelengths of this wave pass a given point on the string every second.