An open organ has a length of 6m. If the speed of sound in air is 340m/s and neglecting the end corrections, the frequency of its first overtone is?
To find the frequency of the first overtone of an open organ pipe, we need to use the formula:
f = (n * v) / (2L)
Where:
f = frequency of the overtone
n = harmonic number (in this case, 2 for the first overtone)
v = speed of sound in air
L = length of the organ pipe
Given the following values:
n = 2 (first overtone)
v = 340 m/s (speed of sound)
L = 6 m (length of the organ pipe)
Plugging in these values into the formula, we get:
f = (2 * 340) / (2 * 6)
Simplifying further:
f = 340 / 6
Calculating this division:
f ≈ 56.67 Hz
Therefore, the frequency of the first overtone of the open organ pipe is approximately 56.67 Hz.