Question

An air column is of length 17 cm. Find the ratio of the frequency of the 5th overtone when the column is closed at one end to that when it is open at both ends. (Speed of sound in air = 340 m/s.)

Hint:

In MHT-CET, if a question asks for a ratio of frequencies of overtones for pipes of the same length, \(v\) and \(L\) always cancel. Don't waste time calculating the numerical values of the frequencies!

Answer 1

Step 1: Understanding the Concept:
Resonance frequencies in air columns depend on whether the ends are open or closed.
- Closed pipe: Only odd harmonics are present (\(f_1, 3f_1, 5f_1, \dots\)).
- Open pipe: All harmonics are present (\(f_1, 2f_1, 3f_1, \dots\)).

Step 2: Key Formula or Approach:
1. For a closed pipe, the frequency of \(n^{th}\) overtone is \((2n + 1) \frac{v}{4L}\).
2. For an open pipe, the frequency of \(n^{th}\) overtone is \((n + 1) \frac{v}{2L}\).

Step 3: Detailed Explanation:
Let \(L = 17\) cm and \(v = 340\) m/s.
1. Closed Pipe (5th overtone, \(n=5\)):
Frequency \(f_{c5} = (2 \times 5 + 1) \frac{v}{4L} = 11 \frac{v}{4L}\).
2. Open Pipe (5th overtone, \(n=5\)):
Frequency \(f_{o5} = (5 + 1) \frac{v}{2L} = 6 \frac{v}{2L}\).
3. Ratio:
\[ \frac{f_{c5}}{f_{o5}} = \frac{11 \frac{v}{4L}}{6 \frac{v}{2L}} = \frac{11}{4} \cdot \frac{2}{6} = \frac{11}{4} \cdot \frac{1}{3} = \frac{11}{12} \] The actual values of \(v\) and \(L\) cancel out, so they do not affect the ratio.

Step 4: Final Answer:
The ratio of frequencies is \(11 : 12\).