SHM Question (1 Viewer)

[Soliah]

The equation of motion of a particle P moving along the line x'ox is given by d^2x/dt^2 = -9x, where x cm is the displacement of P from 0 at time t seconds. Initially, x = 4, dx/dt = -12

i) Prove that its speed in position x is 3 sqrt(32-x^2), and show that it never moves outside a certain interval.

This is what I've done

d/dx(1/2v^2) = - 9x

Integrating both sides

1/2v^2 = -9/2x^2 + C

Initially, x = 4, v = 0

0 = -9/2 x 16 + C
C = 72

Therefore,

1/2v^2 = -9/2x^2 + 72
v^2 = 144 - 9x^2
v = +/- 3 sqrt (16 - x^2)

I don't get 3 sqrt(32-x^2)

What did i do wrong?

 

[ezzy85]

Originally posted by Soliah


Initially, x = 4, v = 0

0 = -9/2 x 16 + C
C = 72

in the question its Initially, x = 4, dx/dt = -12. ie. v = -12. so C should be 144.

 

[Soliah]

:O

Ta.