A body of mass 0.2 kg executes a linear SHM along the X-axis. When it is at position x=0.04 m, its P.E. =0.4 J and K.E. -0.5 J. What is the amplitude of oscillation ?

Options

A. 0.03 m
B. 0.05 m
C. 0.06 m
D. 0.08 m

Answer

C

Explanation

`K.E.=(1)/(2)m omega^(2)(A^(2)-x^(2))`
`P.E. =(1)/(2)m omega^(2)x^(2)`
`therefore (K.E.)/(P.E.)=(A^(2)-x^(2))/(x^(2))=(A^(2))/(x^(2))-1`
`therefore (A^(2))/(x^(2))=1+(K.E.)/(P.E.)`
`therefore (A^(2))/(x^(2))=1 +(0.5)/(0.4)=2.25=(1.5)^(2)`
`therefore (A)/(x)=1.5 therefore A=1.5x=1.5(0.04) therefore A=0.06 m`