Damped simple harmonic motion experiment and questions?

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  • #1
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Homework Statement



Hi, so my question is about a damped simple harmonic motion experiment

The experiment is as follows:

A 30 cm ruler with a needle attached to it is clamped to a bulldog clamp. The needle is placed in a beaker of water so that it is just inside the water ( by about 2cm) and the ruler is displaced so that its starting amplitude is 3.0cm. The ruler is then left to oscillate. The amplitude ( in cm ) after a number 'n' of oscillations is measured. The values used for 'n' are 5,10,15,20,25 and 30. InA is then calculated. A graph of InA against time is drawn.

The questions are:

1. Determine the gradient of the graph

2. Determine the y intercept

3.Using the equation A=A0e^-λn and your values of the gradient and y intercept ,determine the value of λ and A0 ( which should equal the initial amplitude used in the investigation)

4.Justify the number of significant figures used


I don't want to know the actual answers to the question, which I why I haven't uploaded the results or graph for the experiment, I would rather learn the methods behind the questions :)

Homework Equations



A=A0e^-λn
y=mx+c

A0--> initial amplitude

The Attempt at a Solution



1. So the gradient is just Δy/Δx, which equals ΔlnA/Δn, I got a value of -0.0313, however I was unsure what number of decimal places/sig figures I should leave it to, the values in the table are to 1 decimal place, so should it be -0.03?

2. The y intercept. Ok so I was a bit confused here, is the y intercept just simply where it crosses the y axis, so that it is in the form InA= or do I need to calculate A, from InA?
Again same question about number of decimal places/sig figures

3. This question caused me so much difficulty, I honestly have no idea how to approach it. I'm confused as to whether e^-λn= the gradient?! But the gradient is negative, so I can't take ln of both sides?! If I could find λ, then I'm fine with finding A0, its just a case of rearranging?

4. The number of significant figures used in the answer should equal the least number of significant figures used in the data. Since λ is calculated using the gradient, which is to 1 significant figure, then λ must also be to 1 significant figure

Thank you so much :)
 
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Answers and Replies

  • #2
906
88
Are you plotting A vs. t (time) or A vs. n (# of cycles)?

ln(A) = ln(A0e^-λn)
= ln(A0) + ln(e^-λn)
= ln(A0) -λn*ln(e)
= ln(A0) -λn

So ln(A0) = b would be your y-intercept and -λ would be your slope or gradient.

I don't know what table you are referring to but if it really is only to 1 sig fig then the answer can only be given to 1 sig fig. However, you say the displacement A0 is made to 3.0cm which suggests you should be able to measure to 2 sig figs as well.

n is exact so this should be considered as an infinite number of sig figs.
 
  • #3
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I am plotting InA against n :)

Ok so for resolving the equation, did you simply rearrange it into the form y=mx+c and then from my calculated values of the gradient and y intercept I can just state In(A0)= my y intercept and then determine A0 from that? And λ= +gradient ( since my gradient is negative)

Well in the table, there is the value of 'n' number of oscillations, initial A ( A0), final A and InA
I put my values for final A to 1 decimal place e.g 2.4 cm 1.8cm etc, but 2sf, so should InA also be to 2sf? I get confused as to whether the data should be kept to the same number of significant figures or decimal places?

Thank you so much for your reply, I'm actually beginning to understand it! :)
 
  • #4
906
88
Just to be clear, when you type InA you actually mean ln(A) = the natural logarithm of A?

Yes, the y intercept is "c" in your equation y=mx+c. Also, the gradient is negative means m=-λ. You should get a line decreasing in value as you go from left to right with increasing n.

The general rule for taking sig figs of logarithms is that the # of sig figs of the number will be the number decimal places of the logarithm of the number.
 
  • #5
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Yes, sorry my bad. Ahh ok, I understand. So if it is 2.4cm then my ln(A)= 0.056 , for example?
Also I just have one more question In(A) has no units, right? :S
 
  • #6
906
88
ln(2.4) = 0.8756
=0.88 rounded to 2 decimal places

Good question about the units of ln(A). "A" technically needs to be dimensionless so I would assume that here you have divided by the units (cm). So, yes, I would say it has no units.
 
  • #7
13
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Oh god, I didn't calculate what ln(A) was, probably should have...
Oh ok, great!

Ok, thank you for taking the time to help me, I definitely understand it better now!
 
  • #8
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Just wondering one more thing, for my graph, would it be too awkward to put '5, 10' ,15 ,20 ,25,30' every 3 squares for the 'number of oscillations' on the x axis? If I do it every two squares, it looks slightly weird and takes up 3/4 of the page :S
 
  • #9
906
88
There might be a rule about what scale to use for a linear and a ln scale but I am not sure. Usually the people grading these things are extremely anal about how you present these things so you might want to ask them exactly how they want it.

With that said, if it were up to me I would get me some semi-log paper, convert ln(A) to log(A), and show the graph with scale with powers of 10 on the log side and units of 5 on the linear side. But you have to use your own judgement about what makes the graph clear.
 
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