Final answer:
The beat frequency between the fundamental modes of the two organ pipes, one at 10.0°C and the other at 15.0°C, is calculated to be 3 Hz after determining the speed of sound at each temperature and the corresponding frequencies.
Step-by-step explanation:
To calculate the beat frequency between two organ pipes, we need to find the fundamental frequency for each pipe and then subtract one frequency from the other. The fundamental frequency for a pipe open at both ends is given by the formula f = v / (2L) where f is the frequency, v is the speed of sound in air, and L is the length of the pipe. The speed of sound in air can be found using the formula v = v_0 + (0.6 °C^{-1} × T), where v_0 is the speed of sound at 0°C and T is the temperature in °C.
To find the frequencies of the pipes at 10.0°C and 15.0°C, we will first find the respective speeds of sound. For the pipe at 10.0°C: v = 331 m/s + (0.6 * 10) = 337 m/s. The fundamental frequency is then f_1 = 337 m/s / (2 × 0.5m) = 337 Hz.
For the pipe at 15.0°C: v = 331 m/s + (0.6 * 15) = 340 m/s. The fundamental frequency is then f_2 = 340 m/s / (2 × 0.5m) = 340 Hz. The beat frequency, which is the difference between the two frequencies, is f_beat = |340 Hz - 337 Hz| = 3 Hz.
So, the beat frequency between the fundamental modes of the two organ pipes is 3 Hz.