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 ?

Options

A. `sqrt(50) m/s`
B. `sqrt(65) m/s`
C. `sqrt(35) m/s`
D. `sqrt(20)m/s`

Answer

B

Explanation

`v_(1)=8 " at" x, v_(2)=7 " at " x+1 and v_(3)=4 " at " x+2`
In S.H.M., `v=omega sqrt(A^(2)-x^(2)) therefore v^(2)=omega^(2)(A^(2)-x^(2))`
`therefore 64=omega^(2)(A^(2)-x^(2))`
`49= omega^(2)[A^(2)-(x+1)^(2)]`
`16=omega^(2)[A^(2)-(x+2)^(2)]`
`therefore 64-49= omega^(2)[A^(2)-x^(2)-(A^(2)-(x^(2)+2x+1))]`
` therefore 15=omega^(2)[-x^(2)+(x^(2)+2x+1)]=omega^(2)(2x+1)`
and `64-16=omega^(2)[A^(2)-x^(2)-(A^(2)-(x^(2)+4x+4))]`
`48=omega^(2)[-x^(2)+x^(2)+4x+4]=omega^(2)4x+4)`
Dividing (4) by (5), we get
`(15)/(48)=(5)/(16)=(2x+1)/(4x+4)`
`therefore 20x+20=32x+16`
`therefore 12x=4 therefore x=(1)/(3)`
and from (4), `omega^(2)=(15)/(2x+1)=(15)/(2xx(1)/(3)+1)=9`
`therefore omega = +- 3 rad//s`
and from (1), `64=9(A^(2)-(1)/(9))=9A^(2)-1`
`therefore 9A^(2)=65 therefore A^(2)=(65)/(9) therefore A=(sqrt(65))/(3)`
and Maximum velocity `=omega A`
`=3xx(sqrt(65))/(3)=sqrt(65) m//s`