The frequency of a vibrating wire is 200 Hz. The velocity of a particle on the wire is 4.35m*s^(-1) when it is at a distance of half its amplitude. Calculate the acceleration of the particle at that instant.

The frequency of a vibrating wire is 200 Hz. The velocity of a particl

1.	The frequency of a vibrating wire is 200 Hz. The velocity of a particle on the wire is 4.35m*s^(-1) when it is at a distance of half its amplitude. Calculate the acceleration of the particle at that instant.
Answer» Solution :Here, `omega=2pin=2pi200=400pi"rad"s^(-1)" "[becausen=200Hz]` Let A be the AMPLITUDE of the particle. If `X=A/2`, then `v=4.35ms^(-1)` `therefore" "v=omegasqrt(A^(2)-x^(2))` or, `4.35=400pisqrt(A^(2)-A^(2)/4)=400pi(Asqrt3)/2` or, `A=(4.35xx2)/(400pixxsqrt3)m`. `therefore` Required ACCELERATION, `a=omega^(2)x=omega^(2)A/2=(400pi)^(2)xx1/2xx(4.35xx2)/(400pixxsqrt3)` `=(400pixx4.35)/sqrt3=3154ms^(-2)` (approx.)