Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x_(1) is v_(1) and velocity of the particle at position x_(2) is v_(2). Show that the ratio of time period and amplitude is (T)/(A)=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)x_(2)^(2)-v_(2)^(2)-x_(1)^(2)))

Consider a particle undergoing simple harmonic motion. The velocity of

1.	Consider a particle undergoing simple harmonic motion. The velocity of the particle at position x_(1) is v_(1) and velocity of the particle at position x_(2) is v_(2). Show that the ratio of time period and amplitude is (T)/(A)=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)x_(2)^(2)-v_(2)^(2)-x_(1)^(2)))
Answer» Solution :Using equation `v=omegasqrt(A^(2)-x^(2))impliesv^(2)=omega^(2)(A^(2)-x^(2))` Therefore, at position `x_(1), v_(1)^(2)=omega^(2)(A^(2)-x_(1)^(2))""...(1)` Similarly, at position `x_(2), v_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2))""...(2)` SUBTRACTING (2) from (1), we get `v_(1)^(2)-v_(2)^(2)=omega^(2)(A^(2)-x_(1)^(2))-omega^(2)(A^(2)-x_(2)^(2))=omega^(2)(x_(2)^(2)-x_(1)^(2))` `omega=sqrt((v_(1)^(2)-v_(2)^(2))/(x_(2)^(2)-x_(1)^(2)))impliesT=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))` Dividing (1) and (2), we get `(v_(1)^(2))/(v_(2)^(2))=(omega^(2)(A^(2)-x_(1)^(2)))/(omega^(2)(A^(2)-x_(2)^(2)))impliesA=sqrt((v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2))/(v_(1)^(2)-v_(2)^(2)))""...(4)` Dividing equation (3) and equation (4), we have `(T)/(A)=2pisqrt((x_(2)^(2)-x_(1)^(2))/(v_(1)^(2)x_(2)^(2)-v_(2)^(2)x_(1)^(2)))`