The displacement of a particle executing simple harmonic motion is given by `y= A_(0) +A sin omega t+ B cos omega t`. Then the amplitude of its oscillation is given by:

Options

A. `A + B`
B. `A_(0) + sqrt(A^(2) + B^(2))`
C. `sqrt(A^(2) + B^(2))`
D. `sqrt(A_(0)^(2) + (A +B)^(2))`

Answer

B

Explanation

`y= A_(0) + A sin omega t+ B cos omega t`
Suppose `x= A sin omega t + B cos omega t` and taking `A= a cos phi and B= a sin phi`,
`x= a sin omega t cos phi + a cos omega t sin phi`
`=a [sin omega t cos phi + cos omega t sin phi]`
`=a sin (omega t + phi)`
and `A^(2) + B^(2) = a^(2) cos^(2) phi + a^(2) sin^(2) phi`
`=a^(2) [cos^(2) phi + sin^(2) phi]`
`=a^(2)`
`sqrt(A^(2) + B^(2))=a`
`:. y= A_(0) + x = [A_(0) + sqrt(A^(2) + B^(2))] sin (omega t + phi)`
Comparing above equation with `y= A sin (omega t + phi)`
Amplitude `A.= A_(0) + sqrt(A^(2) + B^(2))`