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One end of a wire of radius r is sealedwith the end of another wire of radius 2r . This is used as a sonometer wire with tension T , keeping the junction at the mid-point of the vibrating length . If a stationary wave having a node at the junction is generated,find out the ratio between the number of loops in the two portions . |
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Answer» Solution :Let`rho_(1) and rho_(2) ` be the densities of thematerials of the TWO wires respectively . Mass PER unit length of the FIRST wire , ` m_(1) = pi r^(2) rho _(1)` Mass per unit length of the second wire , ` m_(2) = pi (2r)^(2) rho_(2) = 4 pi r^(2) rho _(2)` `:.(m_(1))/(m_(2)) = (rho_(1))/(4 rho_(2))` The junction is at the mid-point ofthe vibrating length 2L (say) . So , the vibrating length of each wire = l . The two wires arevibrating simultaneoulsy , so the frequency of vibration isthe same . Letit be n .Again, a NODE is generated at the junnction , so an integral number of loops is produced in each wire. if the number of loops in the two wires are p and q , respectively , then ` n = (p)/(2l) sqrt((T)/(m_(1))) and n = (q)/(2l) sqrt((T)/(m_(2)))` `:.(p)/(2l) sqrt((T)/(m_(1))) = (q)/(2l) sqrt((T)/(m_(2)))` or , ` (p)/(q) = sqrt((m_(1))/(m_(2))) = sqrt ((rho_(1))/(4 rho_(2))) = (1)/(2) sqrt((rho_(1))/(rho_(2)))` If the two wires are of the same material, then ` rho_(1) = rho_(2) ` . So , `(p)/(q) = (1)/(2)*` |
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