Answer:
42.5 Hz.
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.
R=2F
As the forces are in same direction so the resultant force will be:
R=F+F
(a)
