Two wires are tied on a sonometer. The tensions, the lengths, the diameters and the densities of the materials of the two wires are in the ratio 8 : 1 , 36 : 35 , 4 : 1 and 1 : 2 , respectively. Find out the number of beats produced per second when the two wires are vibrated simultaneously.Given, the frequency of the wire emitting a tone of higher pitch is 360 Hz .

Two wires are tied on a sonometer. The tensions, the lengths, the diam

1.	Two wires are tied on a sonometer. The tensions, the lengths, the diameters and the densities of the materials of the two wires are in the ratio 8 : 1 , 36 : 35 , 4 : 1 and 1 : 2 , respectively. Find out the number of beats produced per second when the two wires are vibrated simultaneously.Given, the frequency of the wire emitting a tone of higher pitch is 360 Hz .
Answer» Solution :Mass per unit length of wire, `m=(pid^(2))/(4)rho ` , where d= diameter of the wire and `rho` = DENSITY of the material. `:.`Fundamental frequency. `n = (1) /(2L) sqrt((T)/(m)) = (1)/(2l) sqrt((4Y)/(pi d^(2)rho)) = (1)/(ld) sqrt((T)/(pi rho))` So,for the two wires, `(n_(1))/(n_(2)) = (l_(2))/(l_(1))(d_(2))/(d_(1))sqrt((T_(1))/(T_(2))*(rho_(2))/(rho_(1)))=(35)/(36) xx (1)/(4) xx sqrt((8)/(1)xx(2)/(1)) = (35)/(36)` Clearly, `n_(2) gt n_(1)` so the pitch of the tone EMITTED by the 2nd wire is higher . Here, `n_(2)= 360` Hz `:.(n_(1))/(360) = (35)/(36)or , n_(1) = 350 ` Hz ` :.`Number of beats per second = 360 - 350 = 10.