At 16°C, two open-end organ pipes, when sounded together produce 34 beats in 2 seconds. How many beats per second will be produced if the temperature rises to 51°C? Neglect the increase in the length of the pipes.

Question

A pipe open at both ends has a fundamental frequency \(f\) in air. The pipe is now dipped vertically in water drum to half of its length. The fundamental frequency of the air column is now equal to:

Answer 1

To solve this problem, we need to understand the physics behind beats in sound waves. When two sound waves of slightly different frequencies are played together, they produce a phenomenon known as "beats." The beat frequency is the absolute difference between the two sound frequencies.

The formula for beat frequency is given by:

f_{\text{beats}} = |f_1 - f_2|

Given that at 16°C, these organ pipes produce 34 beats in 2 seconds, the beat frequency at this temperature is:

f_{\text{beats}, 16^{\circ}C} = \frac{34}{2} = 17 \text{ beats per second}

In this problem, we assume the length of the pipes doesn't change, so any change in frequency is due entirely to the speed of sound, which depends on temperature.

The speed of sound in air is approximately given by:

v = v_0 \sqrt{\frac{T}{T_0}}

where v_0 is the speed of sound at a reference temperature T_0 (usually taken as 273 K), and T is the current temperature in Kelvin.

Here, we need to find the beat frequency at 51°C. Using the ratio of speeds at different temperatures:

\frac{v_{51^{\circ}C}}{v_{16^{\circ}C}} = \sqrt{\frac{273 + 51}{273 + 16}}

Calculating the above expression gives us:

\sqrt{\frac{324}{289}} \approx 1.0595

This means the frequency at 51°C increases by approximately 5.95% compared to the frequency at 16°C.

The new beat frequency at 51°C is:

f_{\text{beats}, 51^{\circ}C} = 17 \times 1.0595 \approx 18 \text{ beats per second}

Thus, the number of beats per second when the temperature rises to 51°C is 18, which matches option (A).

The correct option is:

(A) 18

Answer 2