A sonometer wire stretched by a load of density `rho` has a fundamental frequency `n_(1)`. When the load is completely immersed in water ( density `rho_("w")`), its fundametnal frequency is `n_(2)`. Show that ` n_(2)/n_(1) = sqrt((rho - rho_("w"))/rho)`.
Two organ pipes , open at both ends, are sounded together and 5 beats are heard per second. The length of the shorter pipe is ` 0.25` m . Find the length of the longer pipe. [Speed of sound in air = 350 m/s, end correction at one end = 0.015 m for both pipes .]

Explanation

On immersing the load in water , the apparent weight of the load ( in water) and hence the tension in the wire decreases because of the force due to buoyancy of water. Since the frequency of the vibrating wire is derectly proportional to the square root of the tension, the frequency decreases.
Let `W_(1)` = weight of the load in air ,
`W_(2)` = (apparent ) weight of the load in water
If `T_(1) and T_(2)` are the respective tensions in the wire , `T_(1) = W_(1) and T_(2) = W_(2)`.
By definition, relative density of a solid
` = ("density of the solid")/("density of water") `
` = ("weight of the body in air")/"(apparent) loss in weight of the body in water )" `
` :. ` Relative density , ` s = rho/rho_("w") = (W_(1))/(W_(1) - W_(2)) = T_(1)/(T_(1) -T_(2) )`
` :. T_(1)/T_(2) = rho/(rho -rho_("w"))`
` n _(1) = 1/(2L) sqrt(T_(1)/m) and n_(2) = 1/(2L) sqrt(T_(2)/m) `
` :. n_(2)/n_(1) = sqrt(T_(2)/T_(1))" " :. n_(2)/n_(1) = sqrt((rho - rho_("w"))/rho)`
which is the required expression.
Data : ` l_(1) = 0.25` m beat frequency = 5 Hz, v = 350 m/s, e = 0.015 m
Since `l_(1) lt l_(2) , n_(1) gt n_(2)`.
` :. n_(1) - n_(2) = 5` Hz
Total end correction for each open pipe, 2e = 0.03 m.
` n_(1) = v/(2(l_(1) +2e)) = (350)/(2(0.25 + 0.03))`
` = 350/(2 xx 0.28)`
` = 350/(0.56) = 625 `Hz
`{:(log 350,," "2.5441),(log 0.56,,ul(-bar(1).7482)),(,,ul(" "2.7959)):}`
AL ` 2.7959 = 625.0`
` :. n_(2) = n_(1) - 5 = 625 - 5 = 620 - 5 = 620 ` Hz
`n_(2) = (v/(2(l_(2)+2e)))`
` :. l_(2) + 2e = v/(2n_(2)) = 350/(2 xx 620 ) = 350/1240 `
` = 0.2823` m
`{:(log 350,," "2.5441),(log 1240,,ul(-3.0934)),(,,ul(" "bar(1).4507)):}`
AL ` bar(1).4507 = 0.2823`
` :. ` The length of the longer pipe,
` l_(2) = 0.2823 - 0.03 = 0.2523` m