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Consider a particle undergoing simple ha...

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)))`

Text Solution

Verified by Experts

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)))`
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