Question

A closed organ pipe 150 cm long gives 7 beats per second with an open organ pipe of length 350 cm, both vibrating in fundamental mode. The velocity of sound is ________ m/s.

Answer 1

Correct Answer: 294

To determine the velocity of sound using the provided information, we will apply the principles of sound waves in organ pipes. As both pipes are in their fundamental modes, we will utilize the fundamental frequency formulas for open and closed pipes.

1. For a closed organ pipe: The fundamental frequency, denoted as \(f_c\), is calculated using the formula \(f_c = \frac{v}{4L_c}\). Here, \(v\) represents the velocity of sound, and \(L_c = 150 \, \text{cm} = 1.5 \, \text{m}\) is the length of the closed pipe.
2. For an open organ pipe: The fundamental frequency, denoted as \(f_o\), is calculated using the formula \(f_o = \frac{v}{2L_o}\). Here, \(L_o = 350 \, \text{cm} = 3.5 \, \text{m}\) is the length of the open pipe.
3. Given that the beat frequency is 7 beats per second: The beat frequency is the absolute difference between the two fundamental frequencies, expressed as \(f_o - f_c = 7 \, \text{Hz}\).
4. The equations for the frequencies are derived as follows: \(f_c = \frac{v}{4 \times 1.5} = \frac{v}{6}\) and \(f_o = \frac{v}{2 \times 3.5} = \frac{v}{7}\).
5. Substituting these into the beat frequency equation yields: \(\left|\frac{v}{7} - \frac{v}{6}\right| = 7\).
6. Solving for \(v\): The equation simplifies to \(\left|\frac{6v - 7v}{42}\right| = 7\), which further reduces to \(\left|\frac{-v}{42}\right| = 7\). This leads to \(\frac{v}{42} = 7\), and consequently, \(v = 42 \times 7 = 294 \, \text{m/s}\).

The computed velocity of sound is \(v = 294 \, \text{m/s}\), which falls within the specified range of (294, 294).