A mass attached to a spring is free to oscillate, with angular velocity `omega`, in a horizontal plane without friction or damping. It is pulled to a distance `x_(0)` and pushed towards the centre with a velocity `v_(0)` at time `t=0`. Determine the amplitude of the resulting oscillations in terms of the parameters `omega, x_(0)and v_(0)`.

Options

A. `sqrt((v_(0)^(2))/(omega^(2))-x_(0)^(2))`
B. `sqrt(omega^(2)v_(0)^(2)+x_(0)^(2))`
C. `sqrt((x_(0)^(2))/(omega^(2))+v_(0)^(2))`
D. `sqrt((v_(0)^(2))/(omega^(2))+x_(0)^(2))`

Answer

D

Explanation

Let the displacement of the block at instant of
time t be, `x=Acos(omegat+phi)`
At `t=0,x=x_(0)`
`thereforex_(0)=Acosphi` . . . (i)
Velocity, `v=(dx)/(dt)=-Aomegasin(omegat+phi)`
At `t=0,v=-v_(0)`
`therefore-v_(0)=-Aomegasinphi`
or `Asinphi=(v_(0))/(omega)`
Squaring and adding (i) and (ii), we get
`A^(2)(sin^(2)phi+cos^(2)phi)=(v_(0)^(2))/(omega^(2))+x_(0)^(2)`
`A=sqrt((v_(0)^(2))/(omega^(2))+x_(0)^(2))`