Question

two particles execute shm with the same time period=6 sec, same amplitude=5cm, have the same mean position. second particle starts after 1 second the first one starts. find the maximum separation bw the two?

Answer 1

@Mashy

Answer 2

omg that is so easy :D

Answer 3

do the simulation in powerpoint.. you ll come to know :D..

Answer 4

hello!??!? what is THE MAX seperation they can have???

Answer 5

lol. we dont have the luxury of powerpoint in exams.

Answer 6

can the time difference be translated into phase difference?

Answer 7

what do you is THE MAX seperation they can have??? think and tell?!.. the max possible??

Answer 8

is it infinity!?? is it like they can have infinite seperation possible??

Answer 9

its shm. so not infinity obv. i cant even think of a start.

Answer 10

forget about start... just draw a case.. where the two particles have max separation!

Answer 11

how can you just draw a case where two particles have max seperation? you have to have an equation and then find a max right?
am i missing out anything?

Answer 12

why can't you think what would be the positions of the two particles having same mean.. and 5cm amplitude.. such that there is max separation.. forget about the initial phase and stuff.. just draw the positions.. such that they have THE MAX separation.. just try answering this question first!

Answer 13

hmm. will it be the position after 1 second when the second particle has to just start?

Answer 14

JUST DRAW the position.. that you think give max serparation!! :O.. draw QUICK!!!

Answer 15

there thats the answer!!.. you cannot have separation more than that!

Answer 16

i think you didnt get the question. or i just missed writing something.
question says they move in the same direction.

Answer 17

wait.. !! i guess too :D

Answer 18

yea you can translate the time difference into phase difference!

Answer 19

how?

Answer 20

consider the initial phase difference to be phi.. then find out what is the additional phase difference they would generate after 1 second.. after that the phase difference remains the same!!

6 seconds corresponds to a phase difference of 2 pi rads
1 second corresponds to there fore 2 pi / 6 rads!

Answer 21

cool. so i write y1=asin(omega t) for the the first particle. y2=asin(wt + pie/3) for the second particle. do y=y1-y2. what next? i thought of differentiating. but differentiate wrt what?

Answer 22

ermm.. wait.. if you do that.. you are assuming at time t=0 they are at same position!..

differentiate with respect to t maybe? :P.. the y is a variable of t only right? then its the t that we need to differentiate.. we ll find the t for max value.. then all you gotta do is substitute that value of t in the equation!

Answer 23

i see. i'll solve it and get back to you.

Answer 24

oki doki!!

Answer 25

got it. thank you! :)

Answer 26

really!?? can you post your work?? cause i didn't get it :D

Answer 27

yea.
y1=5sin(theta)
y2=5sin(theta + pi/3) where theta=wt
y=y2-y1
i get (1/2)sin(theta) + (sqrt3)/2 cos(theta)
differentiate wrt to t and putting equal to 0 i get
tan(pie/3 t)=sqrt3 --> t=1 sec
put this into y and get y=5cm. thats the answer.

Answer 28

wait.. i didn't get how y = 1/2 sin(theta) + sqr.....

i did the differentiation directly!

Answer 29

\[y = Asin(wt) - Asin(wt- \pi/3)\]
differentiating
\[y' = wAcos(wt) - wAcos(wt- \pi/3) = 0\]
and that gives nothing :P

Answer 30

it will. you need to solve cos(wt - pie/3).

Answer 31

wait continuing
\[wAcos(wt) = wAcos(wt- \pi/3) \]
\[\cos(wt) = \cos(wt-\pi/3)\]
hence
\[2n \pi + wt = wt - \pi/3\]

thats what i did.. :D

Answer 32

haha. i dont complicate things. i solved your equation and got the answer. :P

Answer 33

i think ther is a reason m a teacher :D..

Answer 34

haha. i know. teachers like to complicate things.

Answer 35

no .. i meant.. i only solve upto a point.. where the students can reach the final solution :D

Answer 36

oh alright. i get it now. :P

Answer 37

@Mashy is there a way to solve this question using reference circle?

Answer 38

hmm.. should be able to!!..

Answer 39

but it would ultimately be the same thing :P.. you take the projections and stuff!

Answer 40

reference circle method normally is easier and shorter. but i am not able to do it here.

Answer 41

even if you take a reference circle.. you ll have to find the projection of the particle on the diameter.. true? .. so it ultimately would end up wit the same thing :-/

Answer 42

not true. i'll post an image here. just a min.

Answer 43

ok !

Answer 44

not good quality. enough to understand.
but i dont get whats done here.

Answer 45

hmm lets see..

Answer 46

they have converted the time into phase difference.. here .. it would be the difference in angle!!..

so it turns out to be pi/3 rads.. as we had discussed..

however.. they have directly assumed what would be the case where they would have the max separation

Answer 47

this could be one case.. but you see. this does not give max seperation.. m taking projection over the x axis!

Answer 48

but the case they have taken it this on

Answer 49

so basically they shouldnt have taken this directly, right?
and also if particles move in the same direction initially, the figure they have drawn is wrong?

Answer 50

no... its right.. !!