Shm 2

Jul 2008
347
4
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.
 
Jun 2009
806
275
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.
\(\displaystyle a = -n^2*x\)

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

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

\(\displaystyle v*\frac{dv}{dx} = -n^2*x\)
\(\displaystyle \int{vdv} = -n^2\int{xdx}\)

Now find the integration. To find the constant of integration, put v = 0 when x = A, the amplitude.
 
Jul 2008
347
4
Hmmmm i dont get it, am i meant to get v^2 = 300-n^2 x^2 for the integration??? If so, how do i find n???

2. what's 1/2 way between mean position + instantaneous rest?
 
Last edited:
Jun 2009
806
275
Also how exactly do i find the amplitude?
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.
 
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