Question

An object undergoes simple harmonic motion in two mutually perpendicular directions, its position given by r⃗ =Asinωti^+Acosωtj^.

Show that the object remains a fixed distance from the origin (i.e., that its path is circular), and find that distance.

Part B

Find an expression for the object's velocity.

Express your answer in terms of A, ω, t, i^, j^.

Part C

Show that the speed remains constant, and find its value.

Part D

What is the angular speed of the object in its circular path?

A. ω3

B. 2ω

C. ω2

D. ω

E. 3ω

Answer 1

Given position as a function of time of a simple harmonic oscillator is

   r(t) = A Sin Wt i + A cos Wt j

PartA

   say if the oscillator starts from x= 0 m, then r(0) = A sin W*0 i + A cos W*0 j = 0 i + A j

or after one complete rotation w = 2pi /T

   r(T) = A Sin 2piT/T i + A cos 2piT/T j

   r(T) = 0 i + A j

so the oscillator will be at afixed distance form the origin = A

  
Part B

   velocity is rate of change of displacement

  
v(t) = d(r(t))/dt

v(t) = (d/dt)(A Sin Wt i + A cos Wt j)

v(t) = A*w cos Wt i - A W sin Wt j

Part C

speed is = sqrt(A^2W^2 cos^2 Wt +(-A^2W^2 sin^2 Wt))
   = sqrt(A^2W^2 (cos^2 Wt+ sin^2 wt))
   = sqrt(A^2W^2)
   = A*W

PartD

the angular speed of the object in circular path is W