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.

Answer 2

In this problem, we are dealing with resonance in two air columns that are closed at one end. The lengths of these air columns are 100 cm and 120 cm. They both produce sounds in their respective fundamental modes and produce 15 beats per second. We need to find the velocity of sound in these air columns.

To solve this, we start by using the formula for the frequency of a closed-end air column in its fundamental mode:

\(f = \frac{v}{4L}\)

where:

• \(v\) is the velocity of sound in air.
• \(L\) is the length of the air column.

Let the frequencies of the air columns be \(f_1\) and \(f_2\) for lengths 100 cm and 120 cm respectively. Then, we have:

• \(f_1 = \frac{v}{4 \times 100} = \frac{v}{400}\)
• \(f_2 = \frac{v}{4 \times 120} = \frac{v}{480}\)

Given that they produce 15 beats per second, the difference in frequency is:

\(|f_1 - f_2| = 15\)

Substituting the expressions for \(f_1\) and \(f_2\):

\(\left|\frac{v}{400} - \frac{v}{480}\right| = 15\)

By simplifying the difference:

\(\frac{v}{400} - \frac{v}{480} = 15\)

Find a common denominator (2400 in this case) to combine the fractions:

\(\frac{6v}{2400} - \frac{5v}{2400} = 15\)

This simplifies to:

\(\frac{v}{2400} = 15\)

Multiplying through by 2400 gives:

\(v = 15 \times 2400 = 36000 \; \text{cm/s} = 360 \; \text{m/s}\)

Therefore, the velocity of sound in the air columns is 360 m/s. This corresponds to the given option: 360 m/s.