A particle executing a linear SHM has velocities of 8 m/s 7 m/s and 4 m/s, respectively at three points at dis"tan"ces of x m, (x+1) m and (x+2)m from the mean position. What is the maximum velocity of the particle ?

Explanation

Energy at all the three points are equal

`1/2 mv_(1)^(2) + 1/2kx_(2) = 1/2 mv_(2)^(2) + 1/2 k(x+1)^(2)`
`1/2m(64) + 1/2 kx^(2) = 1/2 m(49) + 1/2 k(x+1)^(2)`
`15 = omega^(2) + 2omega^(2)x"….."(i)`
`1/2 mv_(1)^(2) = 1/2 kx^(2) = 1/2 mv_(3)^(2) + 1/2 k (x+ 2)^(2)`
`1/2 m(64) + 1/2 kx^(2) = 1/2m(16) + 1/2 k(x+2)^(2)`
`12 = omega^(2) + omega^(2)x"......."(ii)`
from (i) & (ii) `omega = 3` & `x = 1//3` then, total energy is equal for the maximum kinetic energy
`1/2 m(64) + 1/2 m 1/9 = 1/2 mv_(max)^(2) rArr v_(max) = sqrt(65) m//s`
Or
`v = omegasqrt(a^(2) - x^(2)) rArr 8 = omegasqrt(a^(2) - x^(2))`
`rArr 7 = omegasqrt(a^(2) - (x+1)^(2)), 4 = omegasqrt(a^(2)- (x + 2)^(2))`
After solving the equation
`V_(max) = aomega = sqrt(65) m//s`