posted by srikala on .
The length of a sonometer wire AB is 100cm
where should the two bridges be placed from A to divide the wire in 3 segments whose fundamental frequencies are in the ratio of 1:2:6
Let L1, L2, L3 be the lengths of three segments. Then L1+L2+L3 = 100.
Also from the laws of vibrations of stretched string
f1:f2:f3 = 1:2:6.
L2 =L1/2, L3 =L1/6.
L1 + L1/2 + L1/6 = 100,
10•L1/6 = 100.
L1 = 60cm. L2 = L1/2 = 30 cm, L3 =L1/6 = 10 cm.
The first bridge - L1 = 60 cm
The second bridge – L1+l2 = 90 cm