A particle executing SHM possesses velocities `20cm*s^(-1)and15cm*s^(-1)` at distances 6 cm and 8 cm respectively from its mean position. Calculate the amplitude and the time period of the particle.

Explanation

Velocity of the particle executing SHM,
`v=omegasqrt(A^(2)-x^(2))`
In the first case, `20=omegasqrt(A^(2)-6^(2))" "...(1)`
In the second case, `15=omegasqrt(A^(2)-8^(2))" "...(2)`
Dividing (1) by (2) we get,
`20/15=(omegasqrt(A^(2)-6^(2)))/(omegasqrt(A^(2)-8^(2)))or,4/3=(sqrt(A^(2)-36))/(sqrt(A^(2)-64))or,(A^(2)-36)/(A^(2)-64)=16/9`
or, `16A^(2)-1024=9A^(2)-324`
or, `7A^(2)=700or,A^(2)=100or,A=10cm=0.1m`
From equation (1) we get,
`20=omegasqrt(10^(2)-6^(2))=8omegaor,omega=20/8=5/2"rad"*s^(-1)`
`therefore" "T=(2pi)/omega=(2pi)/5xx2=4/5xx3.14=2.51 s`.