Question

A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)`
A. `2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))`
B. `2pisqrt((upsilon_(1)^(2)+upsilon_(2)^(2))/(x_(1)^(2)+x_(2)^(2)))`
C. `2pisqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(1)^(2)-x_(2)^(2)))`
D. `2pisqrt((x_(1)^(2)+x_(2)^(2))/(upsilon_(1)^(2)+upsilon_(2)^(2)))`

Answer 1

Correct Answer - A
Foa a particle undergoing SHM
`upsilon=omega sqrt(A^(2)-x^(2))`
`:. upsilon_(1)=omegasqrt(A^(2)-x_(1)^(2)) and upsilon_(2)=omegasqrt(A^(2)-x_(2)^(2))`
Squaring we get, `upsilon_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))`
and `upsilon_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))`
Subtracting we get, `upsilon_(1)^(2)-upsilon_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))`
or `omega=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2))) or (2pi)/(T)=sqrt((upsilon_(1)^(2)-upsilon_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))`
`T=2pisqrt((x_(2)^(2)-x_(1)^(2))/(upsilon_(1)^(2)-upsilon_(2)^(2)))`