A string is fixed at both ends to supports 3.50 meters apart and has a linear mass density of 0.00500 kilograms per meter. The string is under a tension of 90.00 newtons. A standing wave is produced on the string with six nodes and five antinodes. Find the wave speed of the standing wave. Find the wavelength of the standing wave. Find the frequency of the standing wave. Find the period of the standing wave.
In this problem statement, we are implicitly told the length of the string of 3.50 meters long, and we’re directly told that the string has a mass density of 0.00500 kilograms per meter. There is a tension force acting along the string of 90.00 newtons. The standing wave produced on the string has six nodes and five antinodes. First, we want to solve for the wave speed, which we’ll call 𝑣. Next we want to solve for the wavelength of the standing wave, 𝜆. We also want to find the wave’s frequency, 𝑓, and the period of the wave, capital 𝑇.
Let’s begin our solution by drawing a diagram of the scenario. We’re told that the standing wave produced has six nodes and five antinodes and that it’s fixed at both ends. In this wave we’ve drawn, there are one, two, three, four, five, six nodes and one, two, three, four, five antinodes. So it matches the description of the standing wave given. We’re given the horizontal span of the standing wave, which we’ve called 𝐿, 3.50 meters. 𝜇 is the linear mass density of the string, and 𝐹 sub 𝑡 is the tension force applied to the string of 90.00 newtons.
To solve for the speed of the standing wave, 𝑣, let’s recall a relationship between 𝑣, tension, and 𝜇, linear mass density. Wave speed, 𝑣, is equal to the square root of the tension on the string divided by its mass density. Applying this relationship to our scenario, we can plug in for the given values of 𝐹 sub 𝑡 and 𝜇. When we calculate this speed, we find a value, to three significant figures, of 134 meters per second. That’s the speed of the standing wave as it moves along the string.
Now let’s look into the wavelength of this standing wave. How long is it? At this point, our diagram becomes very useful to us. We can identify entire wavelengths of the standing wave on our diagram. If we begin at the left side, we can see that there are two full wavelengths and then one half wavelength before the standing wave ends being attached to the right side. We can write that as an equation: two and a half times the wavelength is equal to the overall length 𝐿. So 𝜆 equals 𝐿 over 2.5, or 3.50 meters divided by 2.5 where 2.5 is an exact number. This fraction is equal to 1.40 meters. That’s the linear distance of a complete cycle of the wave.
Now let’s move on to solving for a wave frequency, 𝑓. To do so, let’s recall another relationship for wave speed. The speed of a wave, 𝑣, is equal to the wave frequency, 𝑓, times its wavelength, 𝜆. This means that wave frequency equals wave speed divided by wavelength. Applying this relationship to our scenario, for 𝑣 and 𝜆, we can use the values solved for previously: 134 meters per second divided by 1.40 meters gives us a frequency value of 95.8 hertz. This is how many wave cycles occur each second.
Finally, we want to solve for the period, 𝑇, of the wave. To do this, let’s recall that the period is the inverse of the frequency of the wave. So in our case, 𝑇 equals one over 𝑓, or one over 95.8 hertz. When we enter this value on our calculator, we find a value of 0.0104 seconds. That’s how long it takes for one complete cycle of the wave to pass a point.