A sonometer wire is stretched by a solid mass M. It produces a fundamental note of a certain frequency in tune with a tuning fork when its length is `70` cm. When the mass is immersed in water it is found that the length has to be changed by 5 cm in order to bring it in tune with the same tuning fork. Calculate the relative density of the material of the hanging mass.

Explanation

The fundamental frequency `n_(1)` is given by
`n_(1) = (l)/(2l) sqrt(((T)/(mu)))`
Here l = 70 cm, T = Mg and `mu`, mass per length
`n_(1) = (1)/(2 xx 70)sqrt(((Mg)/(mu))`
When the mass Mis submerged in water, the frequency remains the same for a length of65 cm.
Hence we have
`n_(1) = (1)/(2xx65) sqrt(((M'g)/(mu)))`
Where M' g is the effective weight of mass in water.
From equation-(6.195) and (6.196)
`(1)/(2 xx 70) sqrt(((Mg)/(mu))) = (1)/(2 xx 65) sqrt(((M'g)/(mu)))`
`sqrt(((M)/(M')))` = (14)/(13)
Now using the Archimedes principle, weshall calculate the value of M' in terms of M. Let p be the density of the material of the mass M.
Volume of mass = `(M//p)cm^(3)`
Volume ofwater displaced = `(M/P)cm^(3)`
Upthrust = (M//p)g
Weight of mass M in water `Mg' - ((M)/(P))g`
`M'= M(1-1/p)`
Substituting the value ofM inequation-(6.197), we get
`sqrt(((M)/(M(1-1//p))))` = `(14)/(13)`