SHM Amplitude vs. Frequency graph

[Taniaz]

Homework Statement

A metal hangs vertically from one end of a spring. The other end of the spring is tied to a thread that passes over a pulley and is attached to a vibrator, as shown in the figure. The vibrator is now switched on.

The frequency of vibration is varied from 0.7f to 1.3f where f is the frequency of vibration of the block in the first part. For the block, show the variation with frequency of the amplitude of vibration. Label this line A.

Then same question but now some light feathers are attached to the block to increase air resistance. Draw the new graph.

Homework Equations

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The Attempt at a Solution

I've always known that the frequency doesn't vary with the amplitude but in this case they say it's a curve with its peak at f and the amplitude never reaches 0. I'm not sure why this is the case.

Thanks.

 

[haruspex]

I've always known that the frequency doesn't vary with the amplitude but in this case they say it's a curve with its peak at f and the amplitude never reaches 0. I'm not sure why this is the case.

The independence of frequency and amplitude is for the simple case, no forcing.
When there is a forcing, the forcing itself has an amplitude. The relationship between the frequency of the forcing and the natural frequency determines the ratio between the amplitude of the forcing and the resulting amplitude.

 

[Taniaz]

Ok, I didn't quite understand the last line that you mentioned.
Why does the graph have a peak at f? Is it because that's the natural frequency (so kind of like resonance?)
That's why it's less at 0.7f and 1.3f because it's forced vibration and the natural frequency don't match?
And if I arbitrarily curved it downwards on both sides given that it doesn't meet 0 would be correct?

 

[haruspex]

Ok, I didn't quite understand the last line that you mentioned.
Why does the graph have a peak at f? Is it because that's the natural frequency (so kind of like resonance?)
That's why it's less at 0.7f and 1.3f because it's forced vibration and the natural frequency don't match?
And if I arbitrarily curved it downwards on both sides given that it doesn't meet 0 would be correct?

Yes, that is all correct.
See https://en.m.wikipedia.org/wiki/Resonance#Theory.
Note that there it specifies "lightly damped", rather than undamped, so matches the second part of your question. What will be the key difference when not damped at all?

 

[Taniaz]

In undamped it wouldn't lose any of its energy but in lightly damped it moves almost with the same frequency but loses its energy over time. So in the second case, we will shift the curve downwards since the amplitude of vibration will be lower?

 

[haruspex]

In undamped it wouldn't lose any of its energy but in lightly damped it moves almost with the same frequency but loses its energy over time. So in the second case, we will shift the curve downwards since the amplitude of vibration will be lower?

I was thinking of a rather more dramatic consequence of constantly feeding power into a system with no losses.

 

[Taniaz]

Larger amplitude? Breakage?

 

[haruspex]

Larger amplitude? Breakage?

In the real world, yes, but strictly theoretically what would the graph look like?

 

[Taniaz]

Are you referring to part a of the question or part b?

 

[haruspex]

Are you referring to part a of the question or part b?

The first part, no damping.

 

[Taniaz]

So the graph with the peak at f and graph curving downwards wasn't correct?

 

[haruspex]

So the graph with the peak at f and graph curving downwards wasn't correct?

It is correct, sort of, but how high is the peak?

 

[Taniaz]

The graph they gave us to previously tells us that at f the amplitude is 2

 

[haruspex]

The graph they gave us to previously tells us that at f the amplitude is 2

Fig 4.2 in your attachments has amplitude 2, but that is just showing the natural frequency of the spring/mass system. The vibrator is turned off.
I do not see a graph in your attachments showing what happens with the vibrator turned on. That is for you to draw.

If the are no losses, and energy is constantly fed in from the vibrator (because it is in synch with the natural frequency) what will the energy level be at steady state?

 

[Taniaz]

It will be more but I don't know how much more. So if the amplitude is more, how do we know what it will exactly be with the given information?

 

[haruspex]

It will be more but I don't know how much more. So if the amplitude is more, how do we know what it will exactly be with the given information?

Think a bit harder... energy is constantly being added, at a steady rate, no losses,... how great will the total energy eventually be?