If `f(x) =(x-x_(0)) phi (x) and phi (x)` is continuous at `x=x_(0)` then what is `f(x_(0))` equal to ?

Options

A. `(3T)/(2)`
B. `(T)/(3)`
C. `(2T)/(3)`
D. `(T)/(6)`

Answer

D

Explanation

Given : `x=x_(m)cos(omegat+phi)" At "t=0,x=x_(m)`
`therefore x_(m)=x_(m)cos(omegaxx0+phi)orcosphi=1=cos0^(@)orphi=0^(@)`
`therefore x=x_(m)cosomegat`
When `x=(x_(m))/(2),(x_(m))/(2)=x_(m)cosomegat`
or `cosomegat=(1)/(2)=cos.(pi)/(3)oromegat=(pi)/(3)`
or `t=(pi)/(3omega)=(piT)/(3xx2pi)=(T)/(6)" "(because omega=(2pi)/(T))`