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A sonometer wire of length ℓ has fundamental frequency f0. A bridge is placed at a point Δℓ away from the midpoint of the wire, where Δℓ is much smaller than ℓ. If the two resulting segments are made to vibrate in their fundamental modes, the number of beats produced is—

1. 8f0Δℓ/ℓ
2. f0Δℓ/ℓ
3. 2f0Δℓ/ℓ
4. 4f0Δℓ/ℓ

Correct answer: 8f0Δℓ/ℓ

Solution

Segments l/2 + dl and l/2 - dl give beats f2 - f1 = (v/2)*(2*dl)/((l/2)^2) = 4*v*dl/l^2. With v = 2*f0*l this equals 8*f0*dl/l.

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