What Is the Amplitude and Period in Forced Oscillation?

[~christina~]

[SOLVED] Forced Oscillation Q

Homework Statement

A 2.00kg object attatched to a spring moves without friction and is driven by an external force given by

F= (3.00N)sin(2 \pi t)

The force constant of the spring is 20N/m. Determine

a) period
b) amplitude of motion

Homework Equations

T= 2 \pi / \omega

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

The Attempt at a Solution

a) T= 2 \pi / \omega
T= 2 \pi/ 2 \pi = 1 s

b) Um..A= (F_o /m)/ \sqrt{ (\omega^2- \omega_o^2)^2 + (b \omega /m)^2}

however I'm not sure as to what the

F= (3.00N)sin(2 \pi t)

F_o= 3.00N
\omega = 2 \pi

I'm not sure as to what that omega in the given equation is..is it \omega_o or just the \omega ?

b= 0 so that cancels out...

Thanks alot

 

[rl.bhat]

wo is the angular frequency of the driving force and w is the natural frequency of the oscillating spring. When w = wo the amplitude of the forced oscillation is maximum and that is the condition for the resonance.

 

[~christina~]

wo is the angular frequency of the driving force and w is the natural frequency of the oscillating spring. When w = wo the amplitude of the forced oscillation is maximum and that is the condition for the resonance.

but I was curious to know which one was given in the equation as I know it's written in my book that it is \omega but I don't know which one. And after I know which one it is how do I find the other one since I want to find amplitude.

 

[uart]

wo is the angular frequency of the driving force and w is the natural frequency of the oscillating spring. When w = wo the amplitude of the forced oscillation is maximum and that is the condition for the resonance.

Actually the normal convention is the opposite way around, \omega_0 being the natural frequency.

... but I don't know which one. And after I know which one it is how do I find the other one since I want to find amplitude.

The natural frequency \omega_0 is equal to \sqrt{k/m}, where "k" is the spring constant.

BTW. Obviously you're using a cookie cutter approach of substituting into "Relevant" equations so I'm guessing that a first principle appraoch of solving the systems differential equation is beyond the scope of your current course. I should however point out that the way you are solving this system is fundementally flawed in that with a truly frictionless systems the natural response cannot be ingored for any value of time t. Your "relevent equations" are only finding the particular solution and ingoring the homogenious solution which depends upon the initial conditions.

Actually your approach would however be valid as a good approximation if your system is presented as a near frictionless system in steady state (note that a true frictionless system never actually reaches steady state).

BTW. The truly relevant equation for this system is the DE: m \,\,d^2x/dt^2 + k x= A sin(wt)

 

[~christina~]

b) amplitude of motion

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

however I'm not sure as to

b= 0 so that cancels out... ====> this IS supposed to cancel out right??

I can't figure the correct answer out...

\omega = \omega_o thus they SHOULD cancel out thus = 0
However I said that since the damping was small then shouldn't (frictionless) b= 0 as well?

But if this above is true then the bottom of the Amplitude eqzn would cancel out since it would = 0 + 0?

Then in the end wouldn't the equation be just

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

A= F_o/m

(after cancelling out everything?)
Then
A= 3/2= 1.5 => wrong...

Book answer is 5.09cm

Okay so then I think that I actually plug in for omega's and b is I assume 0 since damping is nonexistant(frictionless)

but after plugging in

\omega= sqrt{ k/m} = sqrt {20.0N/m / 2.00kg}= 3.1623

and \omega = 2 \pi from equation given

and what do I get??

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

A= (3.0N /2.00kg)/ \sqrt{ ((2 \pi )^2- (3.1623))^2 + (0)^2}

A= 1.5/ 5.429 = .27629m => 0.0027629m

Which is ALSO NOT the answer

Answer from book is 5.09cm


Can someone please help me figure out why it isn't the answer the same as the book answer..

THANKS VERY MUCH

 

[rl.bhat]

Natural frequency of the sping is given by wo^2 = k/m = 20/2 = 10
Angular frequency of the driving force =w=2*pi. Substitute these values in the expression for amplitude( put b = 0 ), you will get the answer.

 

[~christina~]

Natural frequency of the sping is given by wo^2 = k/m = 20/2 = 10
Angular frequency of the driving force =w=2*pi. Substitute these values in the expression for amplitude( put b = 0 ), you will get the answer.

I got it THANKS rl.bhat