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Q.

The fundamental frequency of a sonometer wire of length l is n0 . A bridge is now introduced at a distance of Δℓ(<<ℓ) from the centre of the wire. The lengths of wire on the two sides of the bridge are now vibrated in their fundamental modes. Then, the beat frequency nearly is -

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a

n0Δℓ/ℓ

b

8n0Δℓ/ℓ

c

2n0Δℓ/ℓ

d

n0Δℓ/2ℓ

answer is B.

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Detailed Solution

n0=V2l,n1=V2l/2−Δl,n2=V2l/2+Δl​ Beat frequency =n1−n2​=v1l−2Δl−1l+2Δl⇒Beat frequency =n1−n2​=vl+2Δl−l−2Δll2−4Δl2​=v4Δll2−4Δl2=8lΔlv2l=8Δln0l
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