Wave Motion 4 Question 32

30. A string 25cm long and having a mass of 2.5g is under tension. A pipe closed at one end is 40cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats/s are heard. It is observed that decreasing the tension in the string decreases the beat frequency. If the speed of sound in air is 320m/s find the tension in the string.

(1982, 7M)

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Answer:

Correct Answer: 30. (a, b, c, d)

Solution:

  1. By decreasing the tension in the string beat frequency is decreasing, it means frequency of string was greater than frequency of pipe. Thus,

First overtone frequency of string - Fundamental frequency of closed pipe =8

2(v12l1)(v24l2)=8 or v1=l1[8+v24l2]

Substituting the value, we have

v1=0.25[8+3204×0.4]=52m/s Now, v1=TμT=μv12=(ml)v12=(2.5×1030.25)(52)2=27.04N



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