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