SHM question (1 Viewer)

J-Wang · Jul 25, 2012

"A particle is moving in SHM in a straight line. Its maximum speed is 2m/s and its maximum acceleration is 6m/s^2. Find the amplitude and period of motion. "
Anyone have an easy way of doing this?

J-Wang · Jul 25, 2012

does anybody know how to do this?

Sy123 · Jul 25, 2012

Since we are only regarding amplitude and period, then we can fix the Centre of oscillation to be x=0
$\therefore \ddot{x}=-n^2x$
$\frac{1}{2}v^2=\int -n^2 x \ dx \\ \frac{1}{2}v^2=\frac{-n^2 x^2}{2}+C \ \ x=0 \rightarrow v=2 \ \ $Max speed of 2 at C.O.O$ \\ \\ \therefore C=2$
$\frac{1}{2}v^2=\frac{-n^2}{x^2}+2 \\ v^2=-n^2x^2+4 \\ \ddot{x}=6, \ v=0 \ \rightarrow x=a \ \ \ $where a is the amplitude$$
$\therefore n^2a^2=4 ... [1] \\ \\ -6=n^2a ... [2] \\ \\ a=\frac{-2}{3} \ \ \ \therefore $Amplitude is$ \ \ \frac{2}{3}$
$\therefore n=3 \ \ \therefore Period=\frac{2\pi}{3}$

I skipped algebra, pretty much using the logic of SHM in the first place, find 2 equations in terms of n and a, then solve.

SHM question (1 Viewer)

J-Wang

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J-Wang

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Sy123

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