In your physics lab, an oscillator is attached to one end of...

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass $M$ from the free end of the string, producing tension $Mg$ in the string. The oscillator produces transverse waves of frequency $f$ on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities $\mu$. You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary $M$; then you measure the node-to-node distance $d$ for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of $d$. (b) Graph $\mu d^2$ (in kg $\cdot$ m) versus $M$ (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency $f$ of the waves produced on the string by the oscillator. Take $g = 9.80 \, \mathrm{m/s}^2$. (d) For string A ($\mu = 0.0260$ g/cm), what value of $M$ (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of $f$ that you calculated in part (c).

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass $M$ from the free end of the string, producing tension $Mg$ in the string. The oscillator produces transverse waves of frequency $f$ on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities $\mu$. You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary $M$; then you measure the node-to-node distance $d$ for each standingwave pattern and obtain the following data:

(a) Explain why you obtain only certain values of $d$. (b) Graph $\mu d^2$ (in kg $\cdot$ m) versus $M$ (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency $f$ of the waves produced on the string by the oscillator. Take $g = 9.80 \, \mathrm{m/s}^2$. (d) For string A ($\mu = 0.0260$ g/cm), what value of $M$ (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of $f$ that you calculated in part (c).

Transcript

00:01 In our little box here, which is an oscillator, which produces waves on this string, a frequency f.

00:09 Basically, this is attached to a string, which is then attached to a pulley here, fixed to a pulley.

00:14 And hanging from that pulley is a mass m.

00:17 And the distance between the oscillator and the pulley has a length l.

00:21 And basically, we want to give in a node -to -no distance.

00:27 So we're given a node -to -no distance d.

00:29 And recall that the node -to -node distance here, this is a node and this is a node, and this is d, then d is equal to lambda over 2.

00:41 This will be useful.

00:43 So we're given d.

00:44 And we want to explain why we only get certain values of d.

00:49 Well, so if you recall for waves on a string, we have l is equal to n lambda over 2, and that means that lambda here which is equal to 2d is equal to 2l over n so you can cancel these out and that means d is going to be equal to l over n and n here are integers this is the important part 1, 2, 3 dot that dot so the reason we only get certain values for d is because is l is fixed and n is for integers.

01:33 So this is not a continuous function, this is a discrete function because n is integers.

01:37 So that's why we only get certain values for d.

01:42 And so in part b we're asked to graph.

01:45 And what i've done is i've used some plotting software and i have plotted these values here for you to look at.

01:56 And i will bring these onto the screen right now.

02:04 So if we look at the data that we're given in the textbook here, and we plot mu d squared here, which is mu is the mass per unit length, versus m, the mass hanging from the pulley here, we get something, and what's important to see is that it follows, it looks like it follows like a linear trend.

02:28 And we want to explain why the data follows a linear trend, follows a straight line.

02:32 And the reason for this is it follows.

02:36 And that's because if you look at what mu d square is, is going to be equal to mu times lambda over 2 squared, so it's mu landa squared over 4.

02:53 And that's going to be equal through our equation v is equal to root f over mu.

03:00 We can rewrite this as f lambda squared, over 2 v squared.

03:10 And through v equals lambda f, we can write that as we know what the tension is.

03:15 The tension is going to be mg.

03:18 You can write this as mg lambda squared divided by 2, lambda squared, f squared, where here we've used f, the tension is mg, and we've used v as equal to lambda f.

03:34 So these cancel.

03:38 And then what you get is you get m g over 2 f squared...

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