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Two tunig forks A and B produce 6 beats ...

Two tunig forks A and B produce 6 beats `//` sec. On loading A with wax number of beats remains unaltered. However, when a little of wax is removed, the number of beats becomes 2 in 2 seconds. Calculate original frequency of A if that of B is 256.

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Here, frequency of B `=n_(B)=256`
As number of beats `//`sec `=6`
`:. n_(A) =n_(B)+-6=256+-6=262 or 250`.
When A is loaded with wax, its frequency must have decreased from 262 to 250. That is why number of beats `//` sec remains `256-250=6`.
When a little wax is removed, frequency of A increased from 250 to 255. Therefore, number of beats `//` sec `=256-255=1 or 2` beats in 2 second.
Hence original frequency of `A=262Hz. `
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