Calculate the time period of a body executing SHM whose displacement is 0.04 m and 0.05 m and the corresponding velocities are 0.1m/s and `0.08 ms^(-1)`.

Explanation

Given`" "y_(1)=0.04 m and v_(1)=0.1ms^(-1)`
and`" "y_(2)=0.05 m and v^(2)=0.08ms^(-1)`
We know that `v=omega sqrt(A^(2)-y^(2)).`
so that `" "0.1=omega sqrt(A^(2)-(0.04)^(2))`.
`"& "0.08 = omega sqrt(A^(2)-(0.05)^(2))`
Squaring both sides we get
`0.01=omega^(2)(A^(2)-(0.04)^(2))" ...............(1)"`
`0.0064 =omega^(2)(A^(2)-(0.05)^(2))" .................(2)"`
i.e. `(1)-(2), 0.0036=omega^(2)A^(2)-omega^(2)(0.04)^(2)-omega^(2)A^(2)+omega^(2_(0.05)^(2)`
`therefore" "0.0036 =omega^(2)(10^(-4))(5^(2)-4^(2))(5^(2)-4^(2))=9xx106(-4)omega^(2)`
`omega^(2)=(0.0036)/(9xx10^(-4))=(36xx10^(-4))/(9xx10^(-4))=4`
`therefore" "omega=pm 4 " i.e. "omega="4 rad s"^(-2)`
`"i.e, "omega=(2pi)/(T)=4`.
`therefore" "T=(2pi)/(4)=(pi)/(2)=(3.142)/(2)=1.517s`.