11amplitude of SHM
amplitude to hamesha constant hi rehta hai i.e 2mm
correct me if i am rong
a travelling wave y=2mm sin[100pie t-kx] of wavelength lambda=2cm is reflected & transmitted at boundary at x=0 den amplitude of SHM at x=-1 is
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1wat ur then tellin is not amplitude
amplitude to hamesha constant hi rehta hai i.e 2mm
then u r tellin 4/3 must be something else !
it must be ....
1displacement?
y=asin2n(x/lambda +t/T)
FIND ALL VALUES
THEN Y ->@
x=-1
try if this worxxx?
1buttimes is also not given !
then.....either eqn or question or answer is wrong
:)
1no man, amplitude isn't constant after reflection and transmission. There is an expression which relates final amplitudes with initial ones but requires velocity after reflection/transmission. But how do I find that velocity??
1yus u right.i jusrememberedd
Ar =(v2 - v1 /v1 + v2 ) * Ai
n for At =(2 v2 /v1 + v2 ) * Ai
1yeah, but how to find new velocity????
I'm essentially going about in circles [7][7][7]
1can v2 be found by another method ?
boundary at x=0 den amplitude of SHM at x=-1 is ??
mayb new eqn can be made ! and .....
i dono , i m solvin like this 1st time !
1Ar =(v2 - v1 ) * Ai
----------------
(v1 + v2 )
At =(2 v2 ) * Ai
------------
v1 + v2
t -> transmitted
r------------------->reflected
i->>>>>>>>>>>>>>>>>>>>incident
1question is incomplete .. why the hell ar eu all wasting ur time!!
62a travelling wave y=2mm sin[100pie t-kx] of wavelength lambda=2cm is reflected & transmitted at boundary at x=0 den amplitude of SHM at x=-1 is
see the node will be formed at x=0
now try to solve in terms of a standing wave?
1wat is formula ? in wat terms
then do v have to make new exn for 2 waves , tats weast
but wat will b amplitude ?then
how to find velocity aafter trans n refrac ?