1@qwerty
I did not understand you completely.
Can you find the direction of any one wave by using that method so that it will be more clear?
y=Acos(wt-kx)
y=Asin(wt+kx)
Can they form stationary waves? I say yes. My book says no. Need a confirmation.
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22 Answers
11yup i m sorry swordy i misguided u in beginning actually i did not read waves till then properly now i m pretty sure they will form![1]
23wave shape isnt changing with time .
what you are talking is about ∂f/∂t , i.e assuming x =constant, i.e you are talking for a particular x . If you fix x, then f represents diplacement of the element at x as a function of time .
1Btw, df/dt will give you the rate of change of y-co-ordinate of a point on the wave. In a transverse wave every point performs S.H.M. Why will it be 0?
1hmm....dx/dt is coming to be opposite in sign in two cases. So my book is wrong right?
1lol.....read my post again. I am asking you to find out the direction of any one wave
1@talwarmachhli
just now ur friend jeemacchi came and told me that book is correct!
23if f(x,t) is the wave shape , then assuming that it doesnt change, df/dt =df/dx = 0
from either of the two you will get v = dx/dt or v=-dx/dt . That way you can decide the direction
11it clearly represents the standing wave for any wave torepresent a standing wave there shoyld exist fixed points at a given time t at different cordinates this is clearly possible if you sole the cos term.
71You mean there shouldn't be any constant (eg. Î /4) too ?
1combining the 2 eqs ,i m getting y=2Asin(π/4 +ωt)cos(kx-π/4) ;
To represent a standing wave the eq must be of the form ....
y=2Asin(ωt)cos(kx)
so it will not form a stationary wave
1Swordifsh,the waves are of same amplitude,frequency and they are moving in the opposite directions.You can visualise that when they will collide then what will happen.
1y=Asin(wt-kx)
y=-Asin(kx-wt)
You tell me Akhil....are they along the same direction or opposite?
1we can write first wave as y=Asin(90-wt+kx)
Now is the direction of a wave decided by the sign of wt or kx ?