Question

In a resonance experiment, two air columns (closed at one end) of 100 cm and 120 cm long, give 15 beats per second when each one is sounding in the respective fundamental modes. The velocity of sound in the air column is :

• A. 335 m/s
• B. 370 m/s
• C. 340 m/s
• D. 360 m/s

Hint:

For a closed organ pipe, the fundamental frequency is \( f = \frac{v}{4l} \). The beat frequency produced by two sources of sound is the absolute difference of their frequencies. Set up an equation using the given beat frequency and the fundamental frequencies of the two air columns to solve for the velocity of sound \( v \). Ensure consistent units for length (meters in this case).

Answer 1

The Correct Option is D

For a closed-end air column, the fundamental frequency \( f \) is expressed as \( f = \frac{v}{4l} \), where \( v \) denotes the speed of sound and \( l \) represents the air column's length. For an air column of length \( l_1 = 100 \) cm (1 m), the fundamental frequency is \( f_1 = \frac{v}{4l_1} = \frac{v}{4 \times 1} = \frac{v}{4} \) Hz. For a second air column of length \( l_2 = 120 \) cm (1.2 m), the fundamental frequency is \( f_2 = \frac{v}{4l_2} = \frac{v}{4 \times 1.2} = \frac{v}{4.8} \) Hz. The beat frequency, which is the absolute difference between the two frequencies, is given by: \[\text{Beat} = |f_1 - f_2|\] Given a beat frequency of 15 Hz: \[15 = \left| \frac{v}{4} - \frac{v}{4.8} \right|\] \[15 = v \left| \frac{1}{4} - \frac{1}{4.8} \right|\] \[15 = v \left| \frac{4.8 - 4}{4 \times 4.8} \right|\] \[15 = v \left| \frac{0.8}{19.2} \right|\] \[15 = v \left( \frac{8}{192} \right) = v \left( \frac{1}{24} \right)\] Solving for \( v \): \[v = 15 \times 24 = 360 \text{ m/s}\] The speed of sound in the air column is 360 m/s.