3) A mass of 0.02 kg rests on a horizontal table and is...

3) A mass of 0.02 kg rests on a horizontal table and is attached to one end of a spring of spring constant 9 N/m. The other end of the spring is attached to the wall. The damping force on the mass is Fd=-bv, where b=0.035 kg/s and v is the velocity. The mass is subject to a harmonic driving force F= Fo cos(? t) where Fo= 0.15 N and ?= 22 rad/s. a) Determine the amplitude of the oscillation and the phase angle between the driving force and the displacement of the mass for steady state oscillations. b) Write an expression x(t) =......... describing the displacement of the mass as a function of time. Use the numerical value for all constants.

Transcript

00:01 We have a driven harmonic oscillator, so we've got a mass on a spring, it's got a resistive force b and a driving force f, which is f zero cosine omega t.

00:17 We have all those numbers.

00:20 Okay, so the relevant equation is this one, mx double dot plus bx dot plus kx equals f, which is f zero cosine omega t.

00:35 So let's try a solution that looks like this.

00:40 We're expecting to have it be a cosine omega t plus phi, where phi is some phase angle.

00:47 Take the derivative and again, then we'll plug those derivatives back into the equation, x.

01:18 Now i'm going to collect like terms on the left -hand side, because we've got two cosine terms there.

01:38 Okay, then what i'm going to do is i'm going to, we've got the cosine of a sum and the sine of a sum, so i'm going to use our standard trig identities to break that up.

02:22 Okay, then what i'm going to do, so i've got an equation here, i've got some terms that are basically a constant times cosine omega t.

02:37 I have other terms, that are basically a constant times sine omega t, because phi is a constant and omega is a constant and xm is a constant.

02:47 So i'm really just going to separate out the stuff that's cosine omega t from the stuff that's sine omega t.

02:54 I can always do that because they have different time dependents and so because everything else is constant, the coefficients have to separately be equations.

03:08 And what it's going to do is it's going to give me two equations.

03:21 So the first one depends on f0, the second one doesn't.

03:50 And if i look at that second equation, i can cancel out the xm and i can basically solve this, rearrange it, so i've got sine over cosine, which is tangent.

04:06 So the tangent of phi, which like i said is sine over cosine, is just numbers that i already know, so i can calculate it.

04:24 But i want to point one thing out along the way, that there's actually, because i'm going to be really taking an inverse tangent function to get phi, that i have choices i can make and we'll see that it's important that i make those choices.

04:46 We'll talk about that later, but that's a preview of what's going to be happening.

04:52 So i can plug in the numbers and i get a value for the tangent, that's 1 .13.

05:10 And so phi, at least apparently if i just use the calculator, tells me phi is 0 .847 radians...

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