A string vibrates at its third-harmonic frequency. The...

Key Concepts

Standing Waves

Standing waves are the result of the interference of two waves traveling in opposite directions, leading to fixed nodes (points of zero displacement) and antinodes (points of maximum displacement) along a medium such as a string. For a string with fixed ends, the boundary conditions require that the displacement is zero at the endpoints, which in turn determines the spatial pattern of the standing wave.

Harmonics

Harmonics are the natural frequencies at which a system such as a stretched string vibrates. The nth harmonic corresponds to a mode in which the string fits n half-wavelengths between its fixed ends. This quantization results in a sinusoidal mode shape given by y(x) = A sin(n?x/L), where n is an integer, x is the position along the string, L is the string length, and A is the amplitude.

Amplitude Distribution in Standing Waves

The amplitude along a vibrating string in a standing wave pattern is not uniform but varies according to a sine function. Maximum amplitudes (antinodes) occur mid?way between nodes, while the amplitude is zero at the nodes. By knowing the relative amplitude at a specific point on the string, one can use the sine function to relate that point to the overall mode shape and extract information about dimensions such as the string length.

Using Trigonometric Conditions to Determine Physical Quantities

In problems involving standing waves, a measured amplitude at a specific point can be equated to a fractional value of the maximum amplitude using the sine function. Solving the resulting trigonometric equation, often involving the inverse sine function and accounting for the periodicity of sine, can yield unknown physical parameters such as the length of the string.

Transcript

00:01 So i'm going to start by drawing the third harmonic.

00:13 Okay? so if this is zero here, i can see nodes one, two, three, four nodes and three anti -nodes.

00:35 We want to know how long the string is, and we know.

00:41 The amplitude 30 centimeters from one end is half of the maximum amplitude.

00:59 Well, where is this half? where do we reach half of the amplitude? well, i would see half amplitude here and here and here.

01:16 Of course, here, here, and here as well, but that's just a repetition because they're 30 centimeters from the left end or 30 centimeters for the right end.

01:27 Either way, it's 30 centimeters.

01:29 So let's get rid of those three.

01:35 Now, i know that the amplitude at any point is the maximum amplitude times the sign of kx.

01:55 And so maximum amplitude of the sign k is 2 pi over the wavelength times x.

02:19 And we want the amplitude at 30 centimeters from the end to be half the amplitude.

02:28 So, amplitude over a is one -half.

02:40 So therefore, just divide by a on both sides.

02:59 And that needs to be one -half.

03:05 Now, i want to furthermore draw this more complete so that we can see the vibrations up and down here.

03:20 You can't shade in very well with this whiteboard.

03:25 So, where does the sign of something equal one -half? well, that would be when that's something, which is 2 pi over wavelength x, equals either pi over 6, 5 pi over 6, 7 pi over 6, 11 pi over 6, or adding 2 pi to any of those.

04:08 So let's go ahead and solve this.

04:14 We're trying to figure out what x is, but i can tell from observation that x is 1 .5 wavelengths, so a whole wavelength looking at the drawing.

04:36 This is the wavelength here.

04:40 It's one wavelength and a half of a wavelength.

04:43 That would be three halves of the wavelength.

04:46 So let's put this together.

04:48 So we've got, i'm just going to write x here.

04:56 No, actually i'm going to write wavelength is two -thirds of x.

05:06 Wave length is two -thirds of x.

05:09 So two, i'm also going to, eliminate pi and all of these.

05:23 Continue with blue here.

05:26 2 over wavelength.