The pulse shown in Fig. has a speed of `5 cm//s`. If the linear mass of the right string is `0.5` that of the left string , find the ratio of height of the transmitted pulse to that of incident pulse.

Options

A. `(2 sqrt(2))/(( 1 + sqrt(2)))`
B. `(2 sqrt(2))/(( 2 + sqrt(2)))`
C. `(( 2 + sqrt(2)))/(2 sqrt(2))`
D. `(( 1 + sqrt(2)))/(2 sqrt(2))`

Answer

A

Explanation

Since ` v = sqrt(T//mu) , T_(2) = T_(1)`
`mu_(2) = 1//2 mu_(1) , v_(2) = sqrt(2) v_(1)`
`A_(t) = (( 2v_(2))/( v_(1) + v_(2)))A_(i)`
`(A_(t))/( A_(i)) = ( 2 sqrt(2) v_(1))/(v_(1) + sqrt(2) v_(1))`
` = (2 sqrt(2))/(( 1+ sqrt(2)))`