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 an air column closed at one end, the fundamental frequency (first harmonic) is given by: \[ f = \frac{v}{4l} \] where \( v \) is the velocity of sound in the air column and \( l \) is the length of the air column. For the first air column of length \( l_1 = 100 \) cm = 1 m, the fundamental frequency is: \[ f_1 = \frac{v}{4l_1} = \frac{v}{4 \times 1} = \frac{v}{4} \text{ Hz} \] For the second air column of length \( l_2 = 120 \) cm = 1.2 m, the fundamental frequency is: \[ f_2 = \frac{v}{4l_2} = \frac{v}{4 \times 1.2} = \frac{v}{4.8} \text{ Hz} \] The number of beats per second is the absolute difference between the two frequencies: \[ \text{Beat} = |f_1 - f_2| \] Given that the beat frequency is 15 Hz: \[ 15 = \left| \frac{v}{4} - \frac{v}{4.8} \right| \] \[ 15 = v \left| \frac{1}{4} - \frac{1}{4.8} \right| \] \[ 15 = v \left| \frac{4.8 - 4}{4 \times 4.8} \right| \] \[ 15 = v \left| \frac{0.8}{19.2} \right| \] \[ 15 = v \left( \frac{8}{192} \right) = v \left( \frac{1}{24} \right) \] \[ v = 15 \times 24 = 360 \text{ m/s} \] The velocity of sound in the air column is 360 m/s.