InterviewSolution
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InterviewSolution
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A sonometer wire stretched by a load of densityrho has a fundamental frequency n_(1). When the load is completely immersed in water ( density rho_("w")), its fundametnal frequency is n_(2). Show thatn_(2)/n_(1) = sqrt((rho - rho_("w"))/rho). Two organ pipes , open at both ends, are sounded together and 5beats are heard per second. The length of the shorter pipe is0.25 m . Find the length of thelonger pipe. [Speed ofsound in air = 350 m/s, end correction at one end = 0.015 m for both pipes .] |
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Answer» Solution :On immersing the load in water , theapparent weight of the load ( in water) and hence thetension in the WIRE decreases because of theforce due to buoyancy of water. Since the frequency of the vibrating wire is derectly proportional to thesquare root of thetension, thefrequency 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 thewire , `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 |
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