The A string on a cello vibrates in its first normal mode with a frequency of 220 Hz . The vibra

The A string on a cello vibrates in its first normal mode...

Key Concepts

Harmonics in Vibrating Strings

Harmonics are the higher-frequency modes of vibration that occur when the string vibrates in multiple segments. Each successive harmonic corresponds to an integer multiple of the fundamental frequency, reflecting an increase in the number of antinodes and reduction in the wavelength. This concept is crucial to understanding how changes in the mode of vibration affect the frequency.

Fundamental Frequency and Normal Modes

The fundamental frequency is the lowest frequency at which a string vibrates, and it occurs when the string forms a single segment (with nodes at its ends). Normal modes refer to the distinct patterns of standing waves on the string, where each mode corresponds to a harmonic with a specific relationship between wavelength, string length, and frequency.

Wave Equation for Vibrating Strings

This concept describes how transverse waves propagate along a stretched string under tension. It explains that the speed of the wave on the string is determined by the square root of the ratio of the tension to the linear mass density. This relationship forms the core understanding necessary to solve problems involving vibrating strings.

Linear Mass Density

This is the mass per unit length of the string and is a critical parameter in determining the wave speed along the string. It directly influences how the tension in the string translates into vibrational motion and affects the frequencies of the normal modes.

Transcript

00:01 All right, so in this question, we're dealing with a cello string, whose fundamental frequency, f1, is equal to 220 hertz.

00:11 We're also told that its length is 70 centimeters, so we'll go ahead and write down that l is equal to 0 .70 meters.

00:20 The mass of the string is 1 .2 grams, so m equals 1 .20 grams, which is the same as 1 .20 grams, which is the same as 1 .20.

00:32 Times 10 to the negative 3 kilograms.

00:37 And yeah, that is all that we're given here.

00:41 And we are supposed to figure out what the tension in the string is.

00:47 Yeah.

00:49 So the way that we go about this is we got to basically, like think of as we do in other problems, think of the relevant equations and just kind of combine them together, given what we know.

01:01 So the easiest place to start is, just with very simple v equals lambda f so we have f given to us f1 is 220 hertz we don't know v but we know that v is kind of a step on to figuring out what we want to find because we want to find tension and tension is a part of v the wave speed we also don't know lambda lambda has a very simple formula though so lambda is equal to 2l divided by n because we're dealing with a vibrating string.

01:37 But because we're dealing with the fundamental frequency, we know that n is equal to 1, therefore lambda is just 2l.

01:45 And we're given l, so we're all set on figuring out what lambda is.

01:48 Then we need to take apart v a little bit.

01:52 So v is going to be equal to the square root of tension, the tension string, divided by the linear mass density, mu.

02:03 And mu is just equal to the mass of the string divided by the length of the string.

02:08 So we can actually rewrite v as being equal to the square root of tension divided by mass divided by length.

02:18 And now we have everything that we need to know.

02:21 We have, because we are substituting a bunch of things for v, we note the mass of the string and we know the length of the string.

02:30 We don't know t, but that's what we're solving for.

02:32 So that's okay.

02:32 We don't know lambda, but we have a formula for lambda, and that just has l, which we know, and we have f, which we know.

02:38 So we're all set to start plugging things back into the original equation and start doing some algebra to solve for t.

02:45 So if we bring back in our substitutions for v in lambda, we get the equation that the square root of t over m divided by l is equal to 2l times 0.

03:06 So next we need to solve for t.

03:08 So the first thing we do is we square both sides of the equation.

03:11 I'm also going to bring this l up into the numerator since we're, you know, we have, it's already in the denominator and then we're dividing by l again.

03:19 So that's just putting it back into the numerator.

03:22 So we're going to have on the left side, tl divided by m...

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