A particle is executing SHM along a straight line. Its velocities at distances x_(1)andx_(2) from the mean position are V_(1)andV_(2), respectively. Its time period is

A particle is executing SHM along a straight line. Its velocities at d

1.	A particle is executing SHM along a straight line. Its velocities at distances x_(1)andx_(2) from the mean position are V_(1)andV_(2), respectively. Its time period is
Answer» `2pisqrt((x_(1)^(2)+x_(2)^(2))/(V_(1)^(2)+V_(2)^(2)))` `2pisqrt((x_(1)^(2)-x_(2)^(2))/(V_(1)^(2)-V_(2)^(2)))` `2pisqrt((V_(1)^(2)+V_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))` `2pisqrt((V_(1)^(2)-V_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))` SOLUTION :`V_(1)=omegasqrt(a^(2)-x_(1)^(2))or,V_(1)^(2)/OMEGA^(2)=a^(2)-x_(1)^(2)` Similarly, `V_(2)^(2)/omega^(2)=a^(2)-x_(2)^(2)` `therefore" "V_(1)^(2)/omega^(2)-V_(2)^(2)/omega^(2)=x_(2)^(2)-x_(1)^(2)or,omega=sqrt((V_(1)^(2)-V_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))` i.e., `T=(2PI)/omega=2pisqrt((x_(2)^(2)-x_(1)^(2))/(V_(1)^(2)-V_(2)^(2)))`