Is it possible for a body executing linear SHM to have a velocity of 0.03 `ms^(-1)` when its displacement is 0.04 m and a velocity of `0.04 ms^(-1)` , when its displacement is`0.03 m` ? Given amplitude and period of the oscillation are 0.05 m and 6.284 s.
Hint `: v omega sqrt(A^(2)-x^(2))`
or `v^(2) = omega^(2) ( A^(2) - x^(2))`
`0.03^(2) = omega^(2) ( A^(2) - 0.04 ^(2))` ....(1)
`0.04 ^(2) = omega^(2) ( A^(2) - 0.03^(2))` ......(2)
Solve equation (1) and(2) to find a and `omega . T = ( 2pi)/( omega)` . The values of A and T should be same as given values.

Explanation

` v= omegasqrt(A^(2) - x^(2))`
`implies v^(2) = omega^(2)(A^(2) - x^(2))`
`0.03^(2) = omega^(2) ( A^(2) - 0.04^(2) )` ....(i)
`0.04 ^(2)=omega^(2) (A^(2) - 0.03^(2)) ` ....(ii)
Dividing equation(i) by equation (ii), we get
`(9)/(16) = (A^(2)- 16 xx 10^(-4))/(A^(2)- 9 xx 10^(-4))`
`9A^(2) - 81 xx 10^(-4) =16 A^(2) - 256 xx 10^(-4)`
`7A^(2) = 175 xx 10^(-4)`
`A= 5 xx 10^(-2) m = 0.05 m`
Putting this value of A is equation (i), we get `omega = 1 rad s^(-1)`
Time period , `T = ( 2pi)/( omega) = 2pis = 2 xx 3.14 s =6. 284 s`
Values of a and T are same as given values. So the given situations are possible.