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NEET PHYSICSMedium

The number of possible natural oscillations of the air column in a pipe closed at one end of a length of 85 cm85 \text{ cm} whose frequencies lie below 1250 Hz1250 \text{ Hz} is: (velocity of sound 340 ms−1340 \text{ ms}^{-1})

A

4

B

5

C

7

D

6

Step-by-Step Solution

  1. Identify the formula for a closed organ pipe: The natural frequencies for a pipe closed at one end are odd harmonics, given by f=(2n−1)v4Lf = (2n - 1) \frac{v}{4L}, where vv is the speed of sound, LL is the length of the pipe, and n=1,2,3,…n = 1, 2, 3, \dots .
  2. Calculate the fundamental frequency (f1f_1): Given v=340 m/sv = 340 \text{ m/s} and L=85 cm=0.85 mL = 85 \text{ cm} = 0.85 \text{ m}. f1=v4L=3404×0.85=3403.4=100 Hzf_1 = \frac{v}{4L} = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \text{ Hz}
  3. Determine the possible frequencies: A closed pipe produces only odd multiples of the fundamental frequency. The frequencies are: f1=1×100=100 Hzf_1 = 1 \times 100 = 100 \text{ Hz} f2=3×100=300 Hzf_2 = 3 \times 100 = 300 \text{ Hz} f3=5×100=500 Hzf_3 = 5 \times 100 = 500 \text{ Hz} f4=7×100=700 Hzf_4 = 7 \times 100 = 700 \text{ Hz} f5=9×100=900 Hzf_5 = 9 \times 100 = 900 \text{ Hz} f6=11×100=1100 Hzf_6 = 11 \times 100 = 1100 \text{ Hz} f7=13×100=1300 Hzf_7 = 13 \times 100 = 1300 \text{ Hz}
  4. Count frequencies below 1250 Hz1250 \text{ Hz}: The possible natural frequencies below 1250 Hz1250 \text{ Hz} are 100, 300, 500, 700, 900, and 1100 Hz. There are a total of 6 such frequencies.
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