Question

9.
A particle is in simple harmonic motion with period T. At time t = 0 it is halfway
between the equilibrium point and an end poi t of its motion, traveling toward the
end point. The next time it is at the same place is:
A. t =T
B. t =T/2
C. t =T/4
D. t = T/8
E. none of the above

Answer 1

Answer:

This is a fairly good question conceptually from the chapter of simple harmonic motion.

You can for simplicity visualise this happening to a simple pendulum.

That means the Bob is exactly in between the equilibrium position and the extreme position.

That is if Amplitude of the oscillation is A

then at t=0 Position of particle is A/2

Explanation:

Now,

From solving the differential equation for simple harmonic motion of a particle we know that the solution comes out to be

x = A sin(wt+©)

where, w is the angular frequency

and © is the initial phase angle

so at t=0

A/2 = A sin (w(0)+©)

which immediately follow that sin(©) = 1/2 or © = 30°

So,

for the particle to reach it's extreme position it needs to cover A/2 distance more but remember it's velocity is non uniform through out.

Now,

velocity of the given particle

v = Aw cos (WT+30)

At, extreme position, v = 0

so,

cos (WT+30) = 0

or wt + 30 = 90

Wt= 60°

t = π/3w

or it takes the particle to reach the extreme position from it's position at t=0 is π/3w

And thus by symmetry it would again take the same time to get back to that position from equilibrium

And, hence

Total time = 2π/3w

and we know simply that 2π/w is the time period T of the particle

And hence the answer is that total time taken will be T/3.