Shm (1 Viewer)

[theodore223]

a quick question from the 2002 3U exam. Question 4, part c i, for those playing at home:
Particle whose displacement is x, moves in SHM so that
x(double dot) = -16x.
at time t=0, x = 1 and x(dot) = 4.

Show that, for all positions of the particle;
∣x(dot) ∣= 4√(2-x^2)

Hope this makes sence as i suck at writing maths formulas on word or pages.

 

[roadrage75]

rewrite equation

now (v)(dv/dx) = -16x

so vdv = -16xdx

so v^2/2 = -8x^2 + C

when x = 1, v = 4

hence 16/2 = -8 + c

therefore 8 = -8 + c

thus c = 16


hence v^2/2 = -8x^2 + 16

therefore v^2 = -16x^2 + 32

thus abs v = root(32-16x^2)
= 4root(2-x^2)

proven. the reason why its an absolute value is because 4root(2-x^2) cannot be negative.

 

[theodore223]

thanks. makes more sense now. still hate SHM though lol