Help with oscillating spring concept?

In summary: But I think you found the answer to the question that was asked.)In summary, the answer to the question is 640 Hz. This is determined by dividing each frequency option by the given fundamental frequency of 160 Hz and finding the value that results in an integer, which is 4 in this case. This means that 640 Hz is the 4th harmonic of the fundamental frequency.
  • #1
riseofphoenix
295
2
Number1-2.png


Is it acceleration?

I found this graph online:

3bd0c1e6-41bb-4e02-8a67-935333b62d43.gif


Would the answer be acceleration?
 
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  • #2
riseofphoenix said:
Would the answer be acceleration?
Yes, but what's your reasoning?
 
  • #3
Doc Al said:
Yes, but what's your reasoning?

Based on the graph: when v = 0, the mass has no kinetic energy, KE = ½mv2. Therefore, all of its energy is in the form of elastic potential energy, PEelastic = ½kx2. When PEelastic 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.
 
  • #4
Doc Al said:
Yes, but what's your reasoning?

What about this one...number 2:

Number2-1.png


τnet = Iαα = angular acceleration = v/r
I = moment of inertia = mr2

Is the answer center of gravity
 
  • #5
Doc Al said:
Yes, but what's your reasoning?

Because I don't think center of gravity has anything to do with it...
 
  • #6
riseofphoenix said:
Based on the graph: when v = 0, the mass has no kinetic energy, KE = ½mv2. Therefore, all of its energy is in the form of elastic potential energy, PEelastic = ½kx2. When PEelastic 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.
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.
 
  • #7
riseofphoenix said:
What about this one...number 2:

τnet = Iα
This is what you need. If the net torque is constant, what can you say about alpha?
 
  • #8
Doc Al said:
This is what you need. If the net torque is constant, what can you say about alpha?

With respect to I (moment of inertia) you mean?
 
  • #9
riseofphoenix said:
With respect to I (moment of inertia) you mean?
I think we can safely assume that the moment of inertia of the object is constant.
 
  • #10
Doc Al said:
I think we can safely assume that the moment of inertia of the object is constant.

So angular acceleration will not be constant but will be changing.
 
  • #11
riseofphoenix said:
So angular acceleration will not be constant but will be changing.
How did you determine that? Look back at that equation.
 
  • #12
Doc Al said:
How did you determine that? Look back at that equation.

Oh...

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.
 
  • #13
Wait so you're saying if the net torque is constant, then BOTH I (moment of inertia) and alpha (angular acceleration) are constant.
 
  • #14
Conservation of angular momentum has nothing to do with this one.

Imagine if instead of torque, the problem said that there was a constant net force on the object. What would you conclude then?
 
  • #15
riseofphoenix said:
Wait so you're saying if the net torque is constant, then BOTH I (moment of inertia) and alpha (angular acceleration) are constant.
Well we can assume that the object doesn't change its moment of inertia (otherwise the problem is silly). The key conclusion is that alpha is constant (and non-zero). And what does that tell you?
 
  • #16
Doc Al said:
Conservation of angular momentum has nothing to do with this one.

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

Oh!

F = ma... So a constant F means a constant acceleration a. Which means, constant non-zero acceleration.

Velocity is changing?
 
  • #17
riseofphoenix said:
F = ma... So a constant F means a constant acceleration a.
Right.
Which means, if a = 0, then velocity is non-zero.
No, it means that if a = some non-zero value, then velocity is changing.
So in this case, angular velocity would be changing when α (angular acceleration) is constant.
Good!
 
  • #18
Doc Al said:
Right.

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

Good!

So the answer is angular velocity...
:eek:

Thanks for the clarification!
 
  • #19
I have one more quick question!
 
  • #20
Number3-2.png


This may be a really obvious question but...
is it the highest frequency, 740 Hz?
 
  • #21
riseofphoenix said:
This may be a really obvious question but...
is it the highest frequency, 740 Hz?
Once again I must ask: What is your reasoning? (What does it mean to be a harmonic of some fundamental frequency?)
 
  • #22
Doc Al said:
Once again I must ask: What is your reasoning? (What does it mean to be a harmonic of some fundamental frequency?)

Oh wait...this is what I found online

"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
640/160 = 4 Is this the answer?
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:

fn = n • f1
 
Last edited:
  • #23
riseofphoenix said:
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 = 4 Is this the answer?
440/160 = 2.75
Good!

(Nit pick: You're not finding the highest harmonic, just a higher harmonic. The 640 Hz is the only harmonic in the bunch. In this case it's the 4th harmonic.)
 
  • #24
Doc Al said:
Good!

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

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

1. What is an oscillating spring?

An oscillating spring is a mechanical system consisting of a mass attached to a spring, which is then attached to a fixed point. When the mass is displaced from its equilibrium position, it will experience a force from the spring that will cause it to oscillate back and forth.

2. What is the concept of spring constant in oscillating springs?

The spring constant is a measure of the stiffness of a spring. In oscillating springs, it represents the amount of force required to stretch or compress the spring by a certain distance. A higher spring constant means a stiffer spring, and vice versa.

3. How does the mass of an object affect the oscillation of a spring?

The mass of an object attached to a spring affects the period of oscillation, which is the time it takes for the object to complete one full cycle of oscillation. A heavier mass will have a longer period of oscillation compared to a lighter mass on the same spring.

4. What is the equation for the period of oscillation in an oscillating spring?

The equation for the period of oscillation is T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. This means that the period is directly proportional to the square root of the mass and inversely proportional to the square root of the spring constant.

5. How can the amplitude of oscillation be changed in an oscillating spring?

The amplitude of oscillation in an oscillating spring can be changed by altering the initial displacement of the mass or by changing the spring constant. Increasing the initial displacement will result in a larger amplitude, while increasing the spring constant will result in a smaller amplitude.

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