MedVision ad

Circular Motion Question (1 Viewer)

acevipa

Member
Joined
Sep 6, 2007
Messages
238
Gender
Male
HSC
2009
A point p moves in the circle x^2 + y^2 = r^2 with constant angular velocity w. Prove that the foot M of the perpendicular from P onto the x-axis describes a simple harmonic motion about the centre of the circle.

What I've done, is let x = rcos@

Then I differentiate it, I get dx/dt = -rsin@ x (d@/dt)

When I take the second derivative, I get

-rsin@ x (dw/dt) - rcos@ x (d@/dt)^2

But, I'm still having problems proving SHM
 

Trebla

Administrator
Administrator
Joined
Feb 16, 2005
Messages
8,391
Gender
Male
HSC
2006
A point p moves in the circle x^2 + y^2 = r^2 with constant angular velocity w. Prove that the foot M of the perpendicular from P onto the x-axis describes a simple harmonic motion about the centre of the circle.

What I've done, is let x = rcos@

Then I differentiate it, I get dx/dt = -rsin@ x (d@/dt)

When I take the second derivative, I get

-rsin@ x (dw/dt) - rcos@ x (d@/dt)^2

But, I'm still having problems proving SHM
x = rcos θ
dx/dt = - rsin θ.dθ/dt
= - rwsin θ (as dθ/dt = w)
Note that w is constant as given in the question
d²x/dt² = - rw cos θ.dθ/dt
= - rw²cos θ (as dθ/dt = w)
= - w²x
Thus it is simple harmonic
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top