Welcome to NRICH.

 
Simple harmonic motion of two particles
By Carl Evans (P2080) on Sunday, July 16, 2000 - 03:03 pm:

Hi, I'm not quite sure about the following question, and would be grateful if someone could help:

Two particles A and B move in simple harmonic motion about O with amplitude 2×31/2 m and period 2×31/2×p s. A is released from rest at t=0 from the extreme point p where x = 2×31/2. Particle B is released from P at time t = (31/2/2)×p s. Show that the particles will collide (3×31/2/4)×p s after the release of B. Find how far from O the collision will occur.

Thanks
By Sean Hartnoll (Sah40) on Sunday, July 16, 2000 - 04:13 pm:

Write down the equations of motion:

for particle A:

x = M sin(wt + N)

for particle B:

x = M sin(wt + P)

with M,N,P,w constants.

w is found from the period. M is found from the amplitude.

N and P are found from the initial conditions.

Equate the two motions to find information about the collision time/location. I suspect the values have been choosen to make inverting sines and cosines easier.

Sean

By Carl Evans (P2080) on Sunday, July 16, 2000 - 11:46 pm:

Sean,

Perhaps I should have explained the specific part I was having trouble with. I know that I'm supposed to equate the above equations, and when I do, I get to the following stage: sin (t/31/2+ p/2) = sin (t/31/2).

Now, the solution in the back of the book follows on and says:

but sin (t/31/2) = sin (p - t/31/2) or sin (3p - t/31/2).
where t is worked out subsequently.

But I don't understand the above step. I thought it might have been simply, sin(x) is equal to sin (180-x), sin (360+x) etc, but this wouldn't account for the 3p bit.

Why is the above necessary in equating the times when these particles collide?

thanks, Carl

By Sean Hartnoll (Sah40) on Monday, July 17, 2000 - 12:03 am:

Carl,

The best thing to do here is draw a sine curve between 0 and 4p. The results stated from the book are then obvious.

Also from the curve it should also be fairly clear what sort of values you are looking for that satisfy sin(x + p/2) = sin(x). (i.e. can you see from here that this will imply x=p/4 + np, some n integer?)

The results from the book give a method of reaching this answer. Given sin(x+p/2) = sin(x), the first naive thing to try is x = x+p/2, but this obviously doesn't work. What else can we find that is equal to sin(x)?

I don't think the second expression they gave is necessary, as I think the collision will happen within the first period (check this maybe?). But by finding a second expression that is equal to sin(x), we have another oportunity for equating the inside of the sine expression so:

p - x = x + p/2

or

p/2 = 2x

or

x = p/4

which is the result we expected from above, which is good...

put x=t/root(3) to find the time.

Hope this has helped (the main lesson, I think, is to always draw the curves when working with trigonometric functions),

Sean

By Carl Evans (P2080) on Tuesday, July 18, 2000 - 01:44 am:

Sean,

Thanks very much for your help. Indeed, it is obvious once both graphs are drawn where the intersection points are.

The second expression is in fact correct because when the first one is solved for t the answer is root(3)×p/4. This solution is before B is even released, and so it's discarded. However, your formula x=pi/4+np works to produce the correct answer when n=2.

I failed to notice why sin(x+p/2) could not be equivalent to, say, sin(2p+x). I didn't even look at what this really meant for what was in both brackets!

Cheers again for your help

Carl