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A harmonic waves is travelling on string...

A harmonic waves is travelling on string `1`. At a junction with string `2` it is partly reflected and partly transmitted. The linear mass density of the second string is four time that of the first string. And that the boundry between the two strings is at `x = 0`. If tje expression for the incident wave is, `y_(i) = A_(i) cos (k_(1)x - omega_(1)t)`
What are the expression for the transmitted and the reflected waves in terms of `A_(i), k_(1)` and `omega_(1)` ?

Text Solution

Verified by Experts

Since `v = sqrt(T//mu), T_(2) = T_(1)` and `mu_(2) = 4mu_(1)`
we have, `v_(2) = (v_(1))/(2)….(i)`
The frequency does not changes, that is,
`omega_(1) = omega_(2) …..(ii)`
Also, because `k = omega//v`, the wave numbers of the harmonic waves in the two strings are releated by.
`k_(2) = (omega_(2))/(v_(2)) = (omega_(1))/(v_(1)//2) = 2(omega_(1))/(omega_(1)) = 2k_(1) ....(iii)`
The amplitudes are,
`A_(t)=((2v_(2))/(v_(1)+v_(2)))A_(i)=[(2(v_(1)//2))/(v_(1)+(v_(1)//2))]A_(1)=(2)/(3)A_(1)...(iv)`
and `A_(r)=((v_(2)-v_(1))/(v_(1)+v_(2)))A_(i)=[((v_(1//2))-v_(1))/(v_(1)+(v_(1)//2))]A_(i) = (A_(i))/(3)....(v)`
Now with equation `(ii), (iii)` and `(iv)`, the transmitted wave can be written as,
`y_(t) = (2)/(3)A_(i)cos(2k_(1)x - omega_(1)t)`
Similarly the reflected wave can be expressed as,
`= (A_(i))/(3)cos (k_(1)x + omega_(1)t + pi)`
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