Answer:
(a) 391 Hz.
(b) 0.439 m
Step-by-step explanation:
(a)
The fundamental frequency of a closed pipe is given as
fc = v/4l ............ Equation 1
Where fc = fundamental frequency, v = velocity of sound, l = length of pipe
make l the subject of the equation
l = v/4fc ............ Equation 2
Also for an open pipe,
fo = v/2l............ Equation 3
Where fo = fundamental frequency of open pipe.
making l the subject of the equation.
l = v/2fo ............ Equation 4
Therefore, equating equation 4 and equation 2
v/2fo = v/4fc
fo = 2fc ............... Equation 5.
Note: When the cap is removed, the tube becomes an open tube.
Given: fc = 195.5 Hz.
Substitute into equation 5
fo = 2(195.5)
fo = 391 Hz.
(b)
using equation 2 above,
l = v/4fc
Given: v = 343 m/s, fc = 195.5 Hz.
l = 343/(4×195.5)
l = 343/782
l = 0.439 m.
Hence the tube is 0.439 m long.