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A tuning fork produces 10 beats per seco...

A tuning fork produces 10 beats per second with sonometer wire of length 95 cm and 100 cm . Find the frequency of the fork .

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Frequency of the fundamental tone of the sonometer
wire ` n = (1)/(2l) sqrt((T)/(m)),` then
`(n_(1))/(n_(2)) = (100)/(95) = (20)/(19) [here, n_(1) gt n_(2)]`
If the frequency of the tuning fork is x, for 10 beats per second,
` n_(1) - x = 10 and x - n_(2) = 0 `
Adding them, we get `n_(1) - n_(2) = 20`
So, ` (n_(1))/(n_(2)) - 1 = (20)/(n_(2)) "or", (20)/(19) - 1 = (20)/(n_(2)) "or", (1)/(19) = (20)/(n_(2))`
or, `n_(2) = 380 Hz` .
Now , ` x - n_(2) = 10 `
`:. x = 10 + n_(2) = 10 + 380 ` = 390 Hz
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