An organ pipe open at one end has a length of 60 cm. If the speed of sound in air is 340 m/s, what is the lowest frequency produced by the organ pipe?

A) 141.5 Hz

can anybody show me how to do this? not just the answer?

closed end – node

open end - antinode
=> A quarter of the wavelength is on the length of the pipe =>
λ=4L=4•0.6 = 2.4 m
λ=v/f
f=v/ λ=340/2.4 =141.5 Hz

sad

To find the lowest frequency produced by an organ pipe open at one end, we can use the formula:

f = nv/2L

where:
f is the frequency,
n is the harmonic number (for the fundamental frequency, n = 1),
v is the speed of sound in air, and
L is the length of the organ pipe.

In this case, the length of the organ pipe is given as 60 cm, which is equivalent to 0.60 m. The speed of sound in air is given as 340 m/s.

Substituting these values into the formula, we have:

f = (1 * 340) / (2 * 0.60)

f = 340 / 1.20

f = 283.33 Hz

Rounding to the nearest whole number, the lowest frequency produced by the organ pipe is approximately 283 Hz.

Therefore, the correct answer is option B) 283 Hz.