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NEET PHYSICSMedium

The two nearest harmonics of a tube closed at one end and open at the other end are 220 Hz220 \text{ Hz} and 260 Hz260 \text{ Hz}. What is the fundamental frequency of the system?

A

10 Hz10 \text{ Hz}

B

20 Hz20 \text{ Hz}

C

30 Hz30 \text{ Hz}

D

40 Hz40 \text{ Hz}

Step-by-Step Solution

  1. Identify the Properties of the Tube: For a tube closed at one end and open at the other (closed organ pipe), only odd harmonics are present . The frequencies of these harmonics are given by fn=nf1f_n = n f_1, where n=1,3,5,…n = 1, 3, 5, \dots and f1f_1 is the fundamental frequency.
  2. Set up the Equation: Let the two nearest (consecutive) harmonics be nf1n f_1 and (n+2)f1(n+2) f_1. Given: nf1=220 Hzn f_1 = 220 \text{ Hz} (n+2)f1=260 Hz(n+2) f_1 = 260 \text{ Hz}
  3. Calculate the Fundamental Frequency: Subtracting the first equation from the second gives the difference between consecutive harmonics: (n+2)f1−nf1=260−220(n+2) f_1 - n f_1 = 260 - 220 2f1=40  ⟹  f1=20 Hz2 f_1 = 40 \implies f_1 = 20 \text{ Hz} Therefore, the fundamental frequency of the system is 20 Hz20 \text{ Hz}.
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