When a composite wire is made by joining tow wires as shown in figure and possible frequencies of this wire is asked (both ends fixed) then the lowest frequency is that at which individual lowest frequencies of the two wires are equal.
In the figure ,`l_(1)=l_(2)=l` `mu_(1)=(mu_(2))/(9)`=mu. The tension in the strings is T. Here `mu`is the mass per unit length.
What is the lowest frequency such that the junction is an antinode?

Options

A. `1/(4l) sqrt(l/(mu))`
B. `3/(4l)sqrt(l/(mu))`
C. `5/(4l) sqrt(l/(mu))`
D. `7/(4l)sqrt(l/(mu))`

Answer

A

Explanation

For antinode
`f_(1)=((2n_(1)-1))/(4l)sqrt(l/(mu))`
`f_(2)=((2n_(2)-1))/(4l)sqrt(l/(9mu)) =((2n_(2)-1))/(12l)sqrt(l/(mu))`
`f=1/(4l)sqrt(l/(mu)) ((2n_(1)-1))/(4l)=(2n_(2)-1)/(12l)`
`6n_(1)-3 =2n_(2) -1`
`6n_(1)-2n_(2) =2 " " n_(1)=1, n_(2)=2`
`3n_(1)-n_(2)=1`,