Help Simple Harmonic Motion (1 Viewer)

fartman · May 12, 2012

This is Q7 Excercise 25(c) from 3u Fitzpatrick
$Solve the differential equation $ \frac{d^2x}{dt^2}+16x=0 $ subject to the conditions $ x=3 $ and $ \frac{dx}{dt}=16 $ when $ t=0. $ Find the maximum displacement and maximum speed if x metres is the displacement of a particle moving in a straight line at time t seconds.$

How do I find x in terms of t? Any help appreciated

Drongoski · May 12, 2012

fartso - here's my attempt:

$\ddot x = \frac {dx^2}{dt^2} = -16x$

$\therefore \frac {d (\frac {1}{2}v^2)} {dx} = -16x$

$\therefore \int d(\frac {1}{2}v^2) = \int -16x dx$

$\therefore \frac {1}{2}v^2 = -8x^2 + C$

t=0: x = 3 & v = 16

$\therefore \frac {1}{2}(16)^2 = -8 \times 3^2 + C \implies C = 200$

$\therefore 16x^2 = 400 - v^2 \leq 400 \implies |x| \leq 5 m$

$v^2 = 400 - 16x^2 \leq 400 \implies max |v| = $ max speed $ = 20 ms^{-1}$

fartman · May 15, 2012

thanks! your max speed and displacement is right. But for some reason they have x= 4sin(4t) + 3cos(4t) as part of the answer. Is that the solution to the differential equation? How do you get that?

Help Simple Harmonic Motion (1 Viewer)

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