a l 1m long string with a linear mass density of 0003kgm is kept under tension by a force of f 3

Step 1: Calculate the speed of sound in the string using the formula v = sqrt(T/µ), where T is t

A l 1m long string with a linear mass density of 0003kgm is...

A 1m long string with a linear mass density of µ = 0.003kg/m is kept under tension by a force of F = 300N. a. Determine the speed of sound (v = ?) in the string! b. Determine the frequency of the longest wavelength standing wave that can form in the string. c. Another string with the same length and tension, but with a slightly lower linear mass density also emits a sound. The frequency of the beat produced by the fundamental harmonics of the two strings is 5 Hz. Determine the linear mass density of the second string!

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Transcript

00:01 So we have a string that has a length of one meter and a linear mass density of 0 .003 kilograms per meter.

00:12 And we're told that it's kept under a constant tension of 300 newtons.

00:17 And so part a, the first part of this question, asks, what is the speed of sound and the string? so the speed of sound here is going to be the square root of the tension divided by the linear mass density.

00:27 So this will be 300 newtons.

00:31 Divided by 3 times 10 to the negative 3 kilograms per meter.

00:38 And so this will be like basically the square root of 10 ,000.

00:43 And so this is 316 .4 .23 meters per second.

00:51 And then the next part asks, what's the frequency of the longest wavelength standing wave that can form the string? so the longest wavelength is going to be twice the length of the string.

01:02 That's the fundamental mode.

01:04 So, of course, this is two meters.

01:05 So the frequency corresponding to this would be our speed of sound divided by 2l.

01:10 And so this would just be about 158 .11 hertz.

01:19 And then part c says there's another string with the same length and the same tension, but a different mass density.

01:27 And we're told the beat frequencies produced by the fundamental harmonics of the two strings is 5 hertz.

01:32 So like string 1 has a fundamental frequency we just measured...