Final answer:
The frequency of the lowest tone produced by an organ pipe that is open at both ends and is 3.0 meters in length is calculated using the formula f = v / (2L). With a speed of sound of 340 m/s, the frequency is approximately 56.67 Hz, closest to 56 Hz (option A).
Step-by-step explanation:
The question is asking about the frequency of the lowest tone produced by an organ pipe that is 3.0 meters in length and is open at both ends.
To find the frequency (f) of this tone, we use the formula f = v / (2L), where v is the speed of sound and L is the length of the pipe. In this case, v is given as 340 m/s (speed of sound), and the length of the pipe L is 3.0 m.
Using the formula, we get:
f = 340 m/s / (2 × 3.0 m)
f = 340 m/s / 6.0 m
f = 56.67 Hz (approximately)
The frequency of this tone is approximately 56 Hz, closest to choice A.