# Video: Determining Speed and Wavelength of a Wave Propagating on a String

Ed Burdette

The note E₄ is played on a piano and has a frequency of 393.88 Hz. If the linear mass density of this string of the piano is 0.012 kg/m and the string is under a tension of 1000.00 N. What is the speed of the wave on the string? What is the wavelength of the wave?

03:47

### Video Transcript

The note E four is played on a piano and has a frequency of 393.88 hertz. If the linear mass density of this string of the piano is zero point zero one two kilograms per meter and the string is under a tension of 1000.00 newtons, what is the speed of the wave on the string? What is the wavelength of the wave?

In this two-part problem, we’re told the frequency of the wave, 393.88 hertz; we’ll call that 𝑓.

We’re also told the linear mass density of the string, 0.012 kilograms per meter; we’ll refer to that number as the Greek letter 𝜇.

And we’re told the string is under a tension of 1000.00 newtons; that’s a tension force which we’ll call 𝐹 sub 𝑇.

In part one, we want to solve for the speed of the wave on the string; we’ll call that speed 𝑣. And in part two, we want to find the wavelength or 𝜆 of the wave.

Let’s begin with part one, solving for the speed of the wave on the string. To solve for the speed of the wave on the string, let’s recall the relationship between the string tension, the string mass per unit length or 𝜇, and wave speed on the string. Wave speed, 𝑣, is equal to the square root of the tension in the string 𝐹 sub 𝑇 divided by 𝜇, its mass per unit length.

When we apply this relationship to our scenario, note that 𝐹 sub 𝑇, the tension of the string, and 𝜇, the string mass per unit length, are given to us for this part. We can insert these values for have 𝐹 sub 𝑇 and 𝜇 into this equation. Wave speed 𝑣 equals the square root of 1000.00 newtons divided by 0.012 kilograms per meter.

To two significant figures, this equals 290 meters per second That’s how fast the wave moves along this piano string.

Now let’s solve for the length of the wave, 𝜆. To solve for 𝜆, let’s recall the relationship between wave speed, wave frequency, and wavelength.

In general, the speed of a wave 𝑣 is equal to the product of its frequency 𝑓 and its wavelength, 𝜆. When we apply this equation to our situation, we see we can rearrange to solve for 𝜆, the wavelength of the wave.

Let’s divide both sides of the equation by 𝑓. This cancels out the frequency term on the right side of our equation, and we now see that wavelength, 𝜆, is equal to 𝑣, wave speed, divided by 𝑓, wave frequency.

We solved for the wave speed in part one; that’s two 290 meters per second, and we were given the wave frequency 𝑓, 393.88 hertz. We can now rewrite 𝑣 and 𝑓 in terms of the values they have.

The wavelength, 𝜆, is 290 meters per second divided by 393.88 hertz. This fraction is equal to 0.73 meters.

That’s the length of the wave as it travels along the piano wire.