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A stationary wave is given by `y = 5 sin (pi x)/( 3) cos 40 pi t` where `x and y` are in `cm and t` is in second. a. What are the amplitude and velocity of the component wave whose superposition can give rise to this vibration ? b. What is the distance between the nodes ? c. What is the velocity of a particle of the string at the position ` x = 1.5 cm when t = 9//8 s`? |
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Answer» Correct Answer - A::B::C (a) `v= omega/k = (40pi)/(pi//3)` ` = 120 cm//s` (b) Distance between adjacent nodes ` = lambda/2 = pi/k = (pi/(pi//3))` = 3 cm (c) v_P = (dely)/(delt) = (-200pi) sin (pix)/3 sin 40pit` Now, substitute the given value of x and t. |
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