Posted by **srikala** on Tuesday, May 29, 2012 at 12:37am.

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

options:

1.60cm,90cm

2.40cm,80cm

- physics -
**Elena**, Tuesday, May 29, 2012 at 4:19pm
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•L1=f2•L2=f3•L3

Given

f1:f2:f3 = 1:2:6.

Therefore,

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

Answer: 1.60cm,90cm

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