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Driving force- could someone tell me if I'm using the right method for probl

~christina~
Gold Member
712
0
[SOLVED] driving force- could someone tell me if I'm using the right method for probl

1. Homework Statement
a 5.00kg mass attached to a spring of constant 150N/m. A driving force
[tex]F(t)= (120N)cos(\omega_d t) [/tex] is applied to the mass and the damping coefficient (b) is 6.00 N.s/m

a) what is the resonance frequency of this system?

b) what is the amplitude at resonance?

3. The Attempt at a Solution

First, I'm not sure what the dt is for in the equation..

a) what is the resonance frequency of the system?

[tex]F(t)= (120N)cos(\omega_d t) [/tex]

well....I'm not sure about the F(t) since they give it in cos but I assume that I can just write it with cos and it would be the same but not sure about that.. [and what is that d in the equation?]
if I'm right then it would be:

[tex]F(t)= (120N)sin(\omega_d t) [/tex] and

[tex]F_o= 12.0N [/tex]

resonance frequency is simply this I think:

[tex]\omega_o= \sqrt{\frac{k} {m} }[/tex]
and k is given as 150N/m
and the m is 0.500kg right?

b) amplitude at resonance

not sure about this..when something is at resonance what happens?
and what equation would I use to find this?

I thought that I would use this equation but

[tex]A= \frac{F_o/m} {\sqrt{(\omega^2-\omega_o^2)^2 + (\frac{b\omega} {m})^2}}[/tex]

would Fo be the result of pluging in 0 for the time in the driving force equation given?

and is [tex]\omega[/tex] found from this equation=> [tex]\omega= \sqrt{\omega_o^2-(\frac{b} {2m})^2} [/tex]
and then used to plug into the Amplitude of driven oscillator equation to find b?


like the title of the post says, I would really appreciate it if someone could tell
me if I'm doing the problem correctly.


Thank you
 

Answers and Replies

dynamicsolo
Homework Helper
1,648
4
You do not need to use time to answer the questions. (I would suggest you review the meaning of the symbols in the equations you've written, though.)

a) You are correct that the resonant frequency of the system is just the frequency [tex]\omega_o[/tex] at which the "undamped" oscillator would vibrate (the so-called "natural frequency" of the damped oscillator). You can calculate it using the quantities you indicate.

b) For the damped oscillator, the equation

[tex]
A= \frac{F_o/m} {\sqrt{(\omega^2-\omega_o^2)^2 + (\frac{b\omega} {m})^2}}[/tex]

gives the amplitude at which said oscillator will vibrate. If the damped oscillator is being "driven at resonance", that means that the "driving frequency" [tex]\omega_d[/tex] of the external force is equal to the "natural frequency" of the oscillator [tex]\omega_o[/tex].

In the above equation, [tex]F_o[/tex] is the amplitude of the driving force, which is given to you as 120 N. The [tex]\omega[/tex] in that equation is the [tex]\omega_d[/tex] from the driving force equation. But for this question, you will set
[tex]\omega = \omega_d = \omega_o[/tex]. What will become of the first term under the radical in the amplitude equation? As for the other quantities, you are given m and b in the problem statement.

The units in the numerator are N / kg , which equal m/(sec^2). The units in the denominator are rad/(sec^2). So your result will be an amplitude of oscillation in meters.

[The other equation you cited,

[tex]
\omega= \sqrt{\omega_o^2-(\frac{b} {2m})^2} [/tex] ,

is related to the "transient response" of the oscillator and is not used in these questions.]
 
Last edited:
~christina~
Gold Member
712
0
You do not need to use time to answer the questions. (I would suggest you review the meaning of the symbols in the equations you've written, though.)

a) You are correct that the resonant frequency of the system is just the frequency [tex]\omega_o[/tex] at which the "undamped" oscillator would vibrate (the so-called "natural frequency" of the damped oscillator). You can calculate it using the quantities you indicate.

b) For the damped oscillator, the equation

[tex]
A= \frac{F_o/m} {\sqrt{(\omega^2-\omega_o^2)^2 + (\frac{b\omega} {m})^2}}[/tex]

gives the amplitude at which said oscillator will vibrate. If the damped oscillator is being "driven at resonance", that means that the "driving frequency" [tex]\omega_d[/tex] of the external force is equal to the "natural frequency" of the oscillator [tex]\omega_o[/tex].
okay, my book stated that it was "near" natural freqency so I was suspecting that I could say that it was equal for resonance but wasn't sure.

In the above equation, [tex]F_o[/tex] is the amplitude of the driving force, which is given to you as 120 N. The [tex]\omega[/tex] in that equation is the [tex]\omega_d[/tex] from the driving force equation. But for this question, you will set
[tex]\omega = \omega_d = \omega_o[/tex]. What will become of the first term under the radical in the amplitude equation? As for the other quantities, you are given m and b in the problem statement.
well it would equal 0.
The units in the numerator are N / kg , which equal m/(sec^2). The units in the denominator are rad/(sec^2). So your result will be an amplitude of oscillation in meters.

[The other equation you cited,

[tex]
\omega= \sqrt{\omega_o^2-(\frac{b} {2m})^2} [/tex] ,

is related to the "transient response" of the oscillator and is not used in these questions.]
oh..I was confused by this equation. So for damped oscillations I don't use this equation? or is it just when the question refers to driven oscillations?

Thanks dynamicsolo:smile:
 
dynamicsolo
Homework Helper
1,648
4
This equation:

[tex]\omega= \sqrt{\omega_o^2-(\frac{b} {2m})^2} [/tex]

gives you the frequency of the damped oscillator.

When such an oscillator is driven by a periodic external force, there will initially be two simultaneous "responses" added together. The so-called "transient response" is an oscillation which has the same frequency as the damped oscillator, but the amplitude decays exponentially. Together with this is the "steady-state response", which builds up to the amplitude given in this equation:

[tex]A= \frac{F_o/m} {\sqrt{(\omega^2-\omega_o^2)^2 + (\frac{b\omega} {m})^2}}[/tex]

and oscillates at the driving frequency.

So you need to know about both equations. But in an introductory course, we generally discuss that first one when we are talking about a damped oscillator set into motion by release from an initial displacement or by striking it (what is called an "impulsive force"). The second equation comes up in discussion of a damped oscillator which is being acted up by a periodic external force; because the damped frequency is only involved in the transient (short-term) behavior, it is often skipped over, in a first mechanics course, in favor of just talking about the steady-state (long-term) behavior.
 
~christina~
Gold Member
712
0
This equation:

[tex]\omega= \sqrt{\omega_o^2-(\frac{b} {2m})^2} [/tex]

gives you the frequency of the damped oscillator.

When such an oscillator is driven by a periodic external force, there will initially be two simultaneous "responses" added together. The so-called "transient response" is an oscillation which has the same frequency as the damped oscillator, but the amplitude decays exponentially. Together with this is the "steady-state response", which builds up to the amplitude given in this equation:

[tex]A= \frac{F_o/m} {\sqrt{(\omega^2-\omega_o^2)^2 + (\frac{b\omega} {m})^2}}[/tex]

and oscillates at the driving frequency.
okay, I get what your saying.
So you need to know about both equations. But in an introductory course, we generally discuss that first one when we are talking about a damped oscillator set into motion by release from an initial displacement or by striking it (what is called an "impulsive force"). The second equation comes up in discussion of a damped oscillator which is being acted up by a periodic external force; because the damped frequency is only involved in the transient (short-term) behavior, it is often skipped over, in a first mechanics course, in favor of just talking about the steady-state (long-term) behavior.
I'll remember this just in case it comes up on my test tommorow.

Thank you for explaining this dynamicsolo :smile:
 

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