The fundatmental frequency of a sonometer wire increases by `6 Hz` if its tension is increased by `44%` keeping the length constant. Find the change in the fundamental frequency of the sonometer when the length of the wire is increased by `20%` keeping the original tension in the wire.

Answer

5

Explanation

Fundamental frequency of vibrations of string `v=1/(2l)sqrt(T/mu)`
At constant length, `v prop sqrtT ` or `v/sqrtT` = constant `" "...(i)`
At tension is constant, `v prop 1/l` or vl = constant `" "...(ii)`
When tension increased by 44%
New tension, `T. =T + (44)/100 T =1.44T `
If v. is new frequency, then
`(v.)/(sqrtT.)=v/(sqrtT)impliesv.sqrt(((T.)/T))v" "...(iii)`
As `v prop sqrtT` the increase in tension results in an increase in frequency
`:. v.=v+6 `
From equation (iii)
`v+6=sqrt((1.44T)/T)v` or `v+6=1.2v`
`0.2 v=6 ` or `v=6/(0.2)=30Hz`
Hz When length increased by 20%
New length `l.=l+ 200 = 1.21` As vl = constant
Therefore, vl =v. l. `" " v.=l/(l.)v=1/(1.2l)xx30 =25Hz`
Change in fundamental frequency
`Deltav=v.-v = 25-30 = -5 Hz`
Therefore, `Deltav = 5` Hz (decrease)