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A wire of length 25 cm and mass 2 . 5 g ...

A wire of length 25 cm and mass 2 . 5 g is stretched with a fixed tension. The length of a pipe closed at one end is 40 cm . During vibrations, the first overtone of the wire produces 8 beats per second with the fundamental emitted by the pipe . The number of beats reduces with the decrease in tension in the wire. If the velocity of sound in air is 320 ` m*s^(-1)` , find the tension in the wire.

Text Solution

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The fundamental frequency of the pipe closed at one end,
` n_(1) = (V)/(4l) = ((320 xx 100))/(4 xx 40) = 200` Hz
The mass per unit length of the wire,
` m = (2.5)/(25) = 0 . 1 g*cm ^(-1)`
The 1st overtone is the 2nd harmonic of the wire, the frequency is
Clearly, ` n_(2)`
reduces with the decrease in tension T , so ` (n_(2) - n_(1))` also decreases. For this reason, the beat frequency reduces. This means, `n_(2) gt n_(1)` .
As 8 beats are produced per second , we get
` n_(2) - n_(1) = 8 `
or, `n_(2) = n_(1) + 8 = 200 + 8 = 208 ` Hz .
`:. 208 = (sqrt(10T))/(25) or, 10T = ( 208 xx 25)^(2)`
or, ` T = ((208 xx 25)^(2))/(10) = 27 . 04 xx 10 ^(5) dyn = 27 . 0 4 ` N .
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