Forced Oscillator where Damping is Negligible

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Homework Help Overview

The problem involves a forced oscillator with negligible damping, specifically a mass-spring system where a mass of 0.139 kg is attached to a spring with a spring constant of 7.00 N/m. The system is subjected to an oscillating force with an amplitude of 1.88 N, and the goal is to determine the frequency at which the mass will vibrate with an amplitude of 0.430 m.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the simplification of the equation involving the square root and square terms, questioning the algebraic manipulations made in the original poster's calculations.

Discussion Status

There is an ongoing examination of the algebraic steps taken by the original poster, with some participants pointing out potential errors in simplification. The discussion has led to a recognition of mistakes in the calculations, and one participant expresses gratitude for the clarification provided.

Contextual Notes

Participants are working under the assumption that damping is negligible, which simplifies the equations involved. There is a focus on ensuring that the units in the calculations are consistent and correct.

Becca93
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Homework Statement



Damping is negligible for a 0.139 kg mass hanging from a light 7.00 N/m spring. The system is driven by a force oscillating with an amplitude of 1.88 N. At what frequency will the force make the mass vibrate with an amplitude of 0.430 m? There are two possible solutions, enter one of them.


Homework Equations



A = (Fo/m) / (√((ω^2 - ωo^2)^2 + (bω/m)^2)
Damping is negligible, therefore b = 0, therefore
A = (Fo/m) / (√((ω^2 - ωo^2)^2)

ωo = √(k/m)


The Attempt at a Solution



So,
m = 0.139 kg
k = 7.00 N/m
Fo = 1.88 N
A = 0.430 m

A = (Fo/m) / (√((ω^2 - ωo^2)^2)
Rearranged to find ω, is

(ω^2 - √(k/m)^2)^2 = (Fo/m) / A
ω^2 = √((Fo/m)/A) + (k/m)
ω = √( √((Fo/m)/A) + (k/m) )

So,
ω = √( √((1.88/0.139)/0.430) + (7/0.139) )
ω = √( 1.77 + 50.3597)
ω = 7.22 rad/s

ω = 2(pi)f
7.22 /2(pi) = f
f = 1.49 Hz

This is not the correct answer and I have no idea where I'm going wrong.

Am I using the wrong equations? Are my calculations incorrect? Any assistance would be much appreciated.
 
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You simplified ##\sqrt{(\omega^2-\omega_0^2)^2}## incorrectly.
 
vela said:
You simplified ##\sqrt{(\omega^2-\omega_0^2)^2}## incorrectly.

How? I did the following:

√((ω^2 - ωo^2)^2)

Shouldn't the square root and the square cancel, leaving

(ω^2 - ωo^2)

ω^2 - (√k/m)^2

ω^2 - (k/m)

Resulting in:

ω^2 = √((Fo/m)/A) + (k/m)

Where did I go wrong?
 
After setting b=0 and moving some stuff around, you should have
$$A = \frac{F_0/m}{\sqrt{(\omega^2-\omega_0^2)^2}} = \frac{F_0/m}{\omega^2-\omega_0^2}$$My take was that you ended up with
$$A = \frac{F_0/m}{\sqrt{(\omega^2-\omega_0^2)^2}} = \frac{F_0/m}{(\omega^2-\omega_0^2)^2}$$ (compare to "(ω^2 - √(k/m)^2)^2 = (Fo/m) / A" which is what you wrote in your first post) which leads to the incorrect answer.

If you check the units on your result, you'll see they don't work out. That means you messed up the algebra somewhere.
 
vela said:
After setting b=0 and moving some stuff around, you should have
$$A = \frac{F_0/m}{\sqrt{(\omega^2-\omega_0^2)^2}} = \frac{F_0/m}{\omega^2-\omega_0^2}$$My take was that you ended up with
$$A = \frac{F_0/m}{\sqrt{(\omega^2-\omega_0^2)^2}} = \frac{F_0/m}{(\omega^2-\omega_0^2)^2}$$ (compare to "(ω^2 - √(k/m)^2)^2 = (Fo/m) / A" which is what you wrote in your first post) which leads to the incorrect answer.

If you check the units on your result, you'll see they don't work out. That means you messed up the algebra somewhere.

Ahh, yes, I see what I did. I did mess up the algebra without noticing. I had it straight in my head but not on paper.

Thank you very much for pointing that out for me. I've got the correct answer now.
 

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