From the time equations of SHM, prove the relation, `v=+-omegasqrt(A^(2)-x^(2))`

Answer

A

Explanation

Let, `x= A sin omega t`
Then, `sin omega t=(x)/(A)`
Now, `v=(dx)/(dt)=omega Acos omega t`
`=omega Asqrt(1-sin^(2)omega t)`
`=omega Asqrt(1-((x)/(A))^(2))`
`:. v=omega A sqrt((A^(2)-x^(2))/A^(2))`
or `v=+-omega sqrt(A^(2)-x^(2))`