The displacement equation of a particle is `x=3 sin 2t+4cos2t`. The amplitude and maximum velocity will be respectively

Options

A. 5,10
B. 3,2
C. 3,4
D. 4,2

Answer

A

Explanation

Given `x=3sin 2t+4cos2t`
Let, `3=rcostheta and 4=rsin theta` …(i)
then `x=rcosthetasin 2t+rsinthetacos2t`
`=rsin(2t+theta)`
Here, `3^(2)+4^(2)=r^(2)(cos^(theta)+sin^(2)theta)=r^(2)`
or `r=sqrt(3^(2)+4^(2))=5units`
Velocity, `upsilon=(dx)/(dt)=rxx2cos(2t+theta)`
`:. upsilon_(max)=2r=2xx5=10units`