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A string 25 cm long and having a mass of 2.5g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its firsy overtone and the air in the pipe in its fundamental frequency, 8 beats per sec are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320 m/s, find the tension in the string. |
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Answer» Solution :For first overtone (i.E., second harmonic) of thestring `f_(s)=40 sqrt(T)` and for FUNDAMENTAL frequency of closed organ pipe, `f_(c)=200 HZ` So `200-40 sqrt(T)=8 or 40 sqrt(T)-200=8` But decreasing the tension DECREASES the beat frequency. So `40 sqrt(T)=208 i.,e T=[(208)/(40)]^(2)=(5.2)^(2)=27N` |
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