A spring of constant `k=0.5 N//m` and an attached mass m oscillates on a smooth horizontal table. When the mass is at position `x_(1)=0.1m, ` its velocity is `v_(1)=-1ms^(-1)` , and at `x_(2)=0.2m`, it has velocity `v_(2) =0.5ms^(-1)`. Find m and the amplitude A of the motion.

Explanation

The kinetic energy of body attached with spring of spring constant k at any point x is givey by total energy `-` potential energy `=` kinetic energy
i.e., `(1)/(2)kA^(2)-(1)/(2)kx^(2)=(1)/(2)mv^(2)`
where A is the amplitude of oscillation.

As per question,
`(1)/(2)k(A^(2)-x_(1)^(2))=(1)/(2)mv_(1)^(2)`
or `(1)/(2)kA^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)kx_(1)^(2)` ...(i)
and `(1)/(2)kA^(2)=(1)/(2)mv_(2)^(2)+(1)/(2)kx_(2)^(2)` ...(ii)
From (i) and (ii),
`(1)/(2)m(v_(2)^(2)-v_(1)^(2))=(1)/(92)k(x_(1)^(2)-x_(2)^(2))`
or `m(v_(2)^(2)-v_(1)^(2))=k(x_(1)^(2)-x_(2)^(2))`
or `m[(0.5)^(2)-(-1)^(2)]=0.5[0.1)^(2)-(0.2)^(2)]`
or `mxx0.75=0.5xx0.03`
or `m=(0.5xx0.03)/(0.75)=0.02kg`
From (i),
`A=[(mv_(1)^(2)+kx_(1)^(2))/(k)]^(1//2)`
`=[(0.02(-1)^(2)+0.5xx(0.1)^(2))/(0.5)]^(1//2)=0.22m.`