A sonometer wire has a total length of 1m between the fixed ends. Where should the two bridges be placed below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1 : 2 : 3 ?

Options

A. `(1500)/(23)cm,(500)/(23)cm`
B. `(1500)/(23)cm,(300)/(23)cm`
C. `(300)/(23)cm,(1500)/(23)cm`
D. `(1500)/(23)cm,(2000)/(23)cm`

Answer

D

Explanation

Length `(l)=100cm`
Since, frequency `(f)prop(1)/(l)`
`therefore l_(1):l_(2):l_(3)=1:(1)/(3):(1)/(5)=15:5:3`
Now, `15+5+3=23`
`therefore l_(1)=(100)/(23)xx15=(1500)/(23)cm`
`l_(2)=(100)/(23)xx5=(500)/(23)cm`
`l_(3)=(100)/(23)xx3=(300)/(23)cm`
Therefore, the bridges should be placed at `l_(1)` and `(l_(1)+l_(2))cm` i.e. `(1500)/(23)cm and (2000)/(23)cm` from the zero end of the wire.