Acceleration of point x=2m of a transverse wave is 5? m/s^2 (5 m/s^2 in +ve y direction) and slope of waveform is given by dy/dx = 5sin[(pi/12)x] {where pi=22/7}. Then speed of the point at that instant is given in m/s as (?)

Asked by  | 1st Jan, 2013, 12:06: PM

Expert Answer:

(dy/dx )(dx/dt)= 5sin[(pi/12)x](dx/dt)
5?=5sin[(pi/12)x2](dx/dt) at x=2m
d2y/dt2 =acceleration= 5x2?xpi/12cos[(pi/12)x]......differentiating (i)wrt t&putting the value   of (dx/dt)
at x=2m,d2y/dt2 =acceleration=5x2?xpi/12cos[(pi/6)]

Answered by  | 10th Jan, 2013, 05:15: PM

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