A particle executing SHM has velocities `u` and `v`` and acceleration `a` and `b` in two of its position. Find the distance between these two positions.

Options

A. `(u^2-v^2)/(a+b)`
B. `(v^2+u^2)/(a-b)`
C. `(v^2+u^2)/(a+b)`
D. `(v^2-u^2)/(a-b)`

Answer

A

Explanation

Let `x_1` and `x_2` be the distances of the two positions from centre. Then with usual notations
`u^2=omega^2(A^2-x_1^2)` .(i)
`v^2=omega^2(A^2-x_2^2)` .(ii)
`a=omega^2x_1` .(iii)
`b=omega^2x_2` .(iv)
Substracting Eq. (ii) from Eq. (i),
`u^2-v^2=omega^2(x_2^2-x_1^2)` .(v)
Adding Eqs. (iii) and (iv),
`a+b=omega^2(x_1+x_2)` .(vi)
Dividing Eq. (v) by Eq. (vi) `(u^2-v^2)/(a+b)=x_2-x_1`