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i jus need the answer dude.....
the answer i got was 0.71 A, still am checking how i did it
\( y = A sin(\omega t - kx) \) \( F = -ky \) Potential energy = \( 1/2 ky^2\) v = \( dy/dt\) Kinetic Energy = \( 1/2 mv^2 \) equate them
\( 1/2 k A^2 \sin^2(\omega t ) = 1/2 m(\omega\cos(\omega t))^2 \) ... sorry for earlier, there is no kx
thanks fellass............:D
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plus there is A in the right hand term
find the value of t, from above equation ... the omega will remain same, and use that value of t to find the value of y.
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