Amplitude of a Harmonic Oscillator

[Wyes]

Alrighty, so for a maths unit I'm doing at uni I had to form and solve the differential equation of motion for a non-damped forced harmonic oscillator suspended from a ceiling (note we've defined positive x as the downward direction);

mx'' + kx = mg + Fsin(wt), where (units are all SI, so I'm neglecting to mention them) m = 1, k = 400, g = 9.8m/s, F = 20

My result is;

x = -0.0245cos(20t) + w/(w^2 - 400)sin(20t) + 20/(400 - w^2)sin(wt) + 0.0245

This was easy enough, but now I need to find for what frequencies (w) does the extension of the spring never exceed one metre? i.e. |x| <= 1.

Obviously I need this to be true for all t, but I don't know how to find this. I tried just using the sum of the amplitude terms, but my result was a tiny interval which I knew to be incorrect. This makes sense, particularly as the two trig terms with the same frequency will never add 100% constructively.

My problem is now I have no idea how to approach this.

So tell me Escapist, how do I solve this?

 

[Xan174]

I had a crack at this and it appears I came up short. I'll post my thinking just in case it helps anyone else.

My first instinct was to replace all the trig functions with 1, since this is their maximum value. The worst-case scenario (i.e. the maximum amplitude) would occur when all the trig functions were maximised (I know they won't all maximise simultaneously but I've seen reasoning along these lines used on similar problems).

When I did this, the first thing I spotted was that the first term and the last term cancel out, which encouraged me that I was on the right track. I placed the remaining 2 terms over a common denominator (the removal of the time dependence turns this into x is a function of w only) but nothing jumped out at me regarding a solution.

In lieu of an elegant way of doing it I used wolfram alpha to plot the function. It looks like a 1/x graph, but it's asymptote appears at w = -20. For positive w, x is always (considerably) less than 1.

Sorry I couldn't be more help!

 

[nutral]

Don't you have to change the
x = -0.0245cos(20t) + w/(w^2 - 400)sin(20t) + 20/(400 - w^2)sin(wt) + 0.0245
to
1 = -0.0245cos(20t) + w/(w^2 - 400)sin(20t) + 20/(400 - w^2)sin(wt) + 0.0245
And then look at where the limit goes?

 

[Wyes]

I had a crack at this and it appears I came up short. I'll post my thinking just in case it helps anyone else.

My first instinct was to replace all the trig functions with 1, since this is their maximum value. The worst-case scenario (i.e. the maximum amplitude) would occur when all the trig functions were maximised (I know they won't all maximise simultaneously but I've seen reasoning along these lines used on similar problems).

When I did this, the first thing I spotted was that the first term and the last term cancel out, which encouraged me that I was on the right track. I placed the remaining 2 terms over a common denominator (the removal of the time dependence turns this into x is a function of w only) but nothing jumped out at me regarding a solution.

In lieu of an elegant way of doing it I used wolfram alpha to plot the function. It looks like a 1/x graph, but it's asymptote appears at w = -20. For positive w, x is always (considerably) less than 1.

Sorry I couldn't be more help!

This is what I tried. I e-mailed my lecturer asking him if I was approaching the question correctly, and he said that was more or less the way to do it.