An oscillator consists of a block attached to a spring of spring constant K=300 N/m. At some time t the position (measured from its equilibrium position), velocity and acceleration of the block are
`x=0.1 m, v= - 15 m//s " and " a= -90 m//s^(2)`. What is the ampliof motion and the mass of the block ?

Options

A. 0.3 m and 0.3 kg
B. 0.4 m and `(1)/(4)kg`
C. `0.5 " and " (1)/(3) kg`
D. `0.5 m " and " (1)/(4) kg`

Answer

C

Explanation

`because a= - omega^(2)x therefore -90 = omega^(2)(0.1)`
`therefore omega^(2)=900 therefore omega=30//s`
and `v=omegasqrt(A^(2)-x^(2))`
`therefore (v)/(omega)= -(15)/(30)=sqrt(A^(2)-x^(2))`
`therefore (-(1)/(2))^(2)=A^(2)-(0.1)^(2)`
`therefore A^(2)=(1)/(4)+(1)/(100)=(26)/(100) therefore A = (1)/(2)=0.5m`
and `because omega=sqrt((K)/(m)) therefore m=(K)/(omega^(2))=(300)/(900)=(1)/(3)kg`