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Is it acceleration?

I found this graph online:

Would the answer be acceleration?

- Thread starter riseofphoenix
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- #1

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Is it acceleration?

I found this graph online:

Would the answer be acceleration?

- #2

Doc Al

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Yes, but what's your reasoning?Would the answer be acceleration?

- #3

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Based on the graph: when v = 0, the mass has no kinetic energy, KE = ½mvYes, but what's your reasoning?

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What about this one...number 2:Yes, but what's your reasoning?

τ

α = angular acceleration = v/r

I = moment of inertia = mr

Is the answer

- #5

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Because I don't think center of gravity has anything to do with it...Yes, but what's your reasoning?

- #6

Doc Al

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Good. Another way to look at it is in terms of Hooke's law. The restoring force--and thus the acceleration--is maximum when the displacement from equilibrium is maximum.Based on the graph: when v = 0, the mass has no kinetic energy, KE = ½mv^{2}. Therefore, all of its energy is in the form of elastic potential energy, PE_{elastic}= ½kx^{2}. When PE_{elastic}is maximum, the restoring force within the spring is also maximized. This results in the mass' acceleration to be maximized as the spring acts to return the mass to its equilibrium position.

- #7

Doc Al

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This is what you need. If the net torque is constant, what can you say about alpha?What about this one...number 2:

τ_{net}= Iα

- #8

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With respect to I (moment of inertia) you mean?This is what you need. If the net torque is constant, what can you say about alpha?

- #9

Doc Al

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I think we can safely assume that the moment of inertia of the object is constant.With respect to I (moment of inertia) you mean?

- #10

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So angular acceleration will not be constant but will be changing.I think we can safely assume that the moment of inertia of the object is constant.

- #11

Doc Al

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How did you determine that? Look back at that equation.So angular acceleration will not be constant but will be changing.

- #12

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Oh...How did you determine that? Look back at that equation.

I'm not sure. I guessed...

Cause I remember for the Conservation of Angular Momentum concept,

L = Iω

If L is to stay constant, then when you increase I, ω decreases. And if you decrease I, ω increases.

I thought it was a similar concept with this question - the whole inverse relationship.

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- #14

Doc Al

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Imagine if instead of torque, the problem said that there was a constant net force on the object. What would you conclude then?

- #15

Doc Al

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Well we can

- #16

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Oh!

Imagine if instead of torque, the problem said that there was a constant net force on the object. What would you conclude then?

F = ma.... So a constant F means a

Velocity is changing?

- #17

Doc Al

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Right.F = ma.... So a constant F means aconstantacceleration a.

No, it means that if a = some non-zero value, then velocity isWhich means, if a = 0, thenis non-zero.velocity

Good!So in this case, angularvelocitywould be changing when α (angular acceleration) is constant.

- #18

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So the answer is angular velocity...Right.

No, it means that if a = some non-zero value, then velocity ischanging.

Good!

Thanks for the clarification!

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I have one more quick question!

- #20

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This may be a really obvious question but...

is it the highest frequency, 740 Hz?

- #21

Doc Al

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Once again I must ask: What is your reasoning? (What does it mean to be a harmonic of some fundamental frequency?)This may be a really obvious question but...

is it the highest frequency, 740 Hz?

- #22

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Oh wait...this is what I found onlineOnce again I must ask: What is your reasoning? (What does it mean to be a harmonic of some fundamental frequency?)

"A 'harmonic' is defined as an integer multiple of a the fundamental frequency F. Harmonics of F will have frequencies NF where N is an integer. The case where N = 1 is the fundamental frequency itself.

So in the example we are given, we should look at the frequencies 100, 200, 250, etc and try to find the highest value of F for which all of the given frequencies are integer multiples. Clearly this is 50Hz:-

100 = 2*50 (second harmonic of 50Hz)

200 = 4*50 (4th harmonic)

250 = 5*50 (5th harmonic)

300 = 6*50 (6th harmonic)

In principle it is possible that the fundamental could be 25Hz or 5Hz or even 2Hz, but if any of these were the case we would expect that there would be harmonics such as 275Hz (11*25) or 280 (56*5), etc, but we are told that these are not observed.

It seems likely therefore that the fundamental is 50Hz"

So essentially, to find the harmonics (using the fundamental frequency of 160 Hz), all I have to do is divide each option by 160 to see if it gives me an integer.

540/160 = 3.375

740/160 = 4.625

440/160 = 2.75

Note to self:

The frequency (f) of the nth harmonic (where n represents the harmonic # of any of the harmonics) is n times the frequency of the first harmonic. In equation form, this can be written as:

f

Last edited:

- #23

Doc Al

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Good!So essentially, to find the highest harmonic (using the fundamental frequency of 160 Hz), all I have to do is divide each option by 160 to see if it gives me an integer.

540/160 = 3.375

740/160 = 4.625

640/160 = 4Is this the answer?

440/160 = 2.75

(Nit pick: You're not finding the

- #24

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Sorry I took so long to respond I was trying to finish dinner as fast as i could...Good!

(Nit pick: You're not finding thehighestharmonic, just a higher harmonic. The 640 Hz is the only harmonic in the bunch. In this case it's the 4th harmonic.)

And ok - so the answer is 640 Hz (even though the 4th harmonic is the only harmonic in the choices I'm given)

Thanks for taking the time to answer my questions!

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