A particle moves in SHM with centre O and passes through O with speed 10sqrt(3) cm/s. by integrating acceleration = - n^2 * x, calculate the speed when the particle is halfway between its mean position and a point of instantaneous rest.

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- May 10th 2010, 03:41 AM #1

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- May 10th 2010, 05:12 AM #2

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$\displaystyle a = -n^2*x$

$\displaystyle \frac{dv}{dt} = -n^2*x$

$\displaystyle \frac{dv}{dx}*\frac{dx}{dt} = -n^2*x$

$\displaystyle v*\frac{dv}{dx} = -n^2*x$

$\displaystyle \int{vdv} = -n^2\int{xdx}$

Now find the integration. To find the constant of integration, put v = 0 when x = A, the amplitude.

- May 13th 2010, 04:24 AM #3

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- May 13th 2010, 04:40 AM #4

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- May 13th 2010, 05:15 AM #5

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When you do the integration, you get

v^2/2 = - n^2*x^2/2 + C or

v^2 = -n^2*x^2 + K.

When x = 0, v = 10sqrt(3), v^2 = 300.

v^2 = -n^2*(x^2) + 300.

When x = A, the amplitude, v = 0. So

0 = -n^2A^2 + 300.

n^2*A^2 = 300.

Velocity at A/2 is given by

v^2 = -n^2*(A/2)^2 + 300

Substitute the value for n^2A^2 and find v.