Question

The two parts of a sonometer wire divided by a movable knife edge , differ in length by `2 mm` and produce `1 beat//s` , when sounded together . Find their frequencies if the whole length of wire is `1.00 m`.

Answer 1

Let `l_(1)` and `l_(2)` be the lengths of the two parts of a sonometer wire divided by a movable knife edge.
`:. l_(1)+l_(2)=100cm` ,brgt and `l_(1)-l_(2)2mm =0.2cm`
Adding and subtracting, we get,
`l_(1)=50.1cms and l_(2)=49.9cms`
Let, `v_(1)` and `v_(2)` be the frequencies of the sonometer wire corresponding to lengths `l_(1)` and `l_(2)` respectively.
Number of beats produced `//` sec. `=v_(2)-v_(1)=1` ...(i)
According to the law of length,
`(v_(2))/(v_(1))=(l_(1))/(l_(2))=(50.1)/(49.9)or v_(2)=(50.1)/(49.9)v_(1)`
Putting in (i), we get
`(50.1)/(49.9)v_(1)-v_(1)=1 or (50.1v_(1)=49.9v_(1))/(49.9)=1`
or `0.2v_(1)=49.9`
or `v_(1)=(49.9)/(0.2)=249.5Hz`
From (i), we have `v_(2)=v_(1)+1=249.5+1`
`=250.5Hz`