Question

What is the lowest frequency that will resonate in an organ pipe 2.00 m in length, closed at one end? The speed of sound in air is 340 m/s.

Answer 1

42.5 Hz.

Step-by-step explanation:

The fundamental frequency of a closed pipe is given as

f₀ = v/4l....................... Equation 1

Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe

Given: v = 340 m/s, l = 2.00 m.

Substitute into equation 1

f₀ = 340/(4×2)

f₀ = 340/8

f₀ = 42.5 Hz.

Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.