When a metal pipe is cut into two pieces, the lowest resonance frequency in one piece is 244Hz and that for the other is 454Hz. What resonant frequency would have been produced by the original length of pipe? (Assume the speed of sound in air is 345 m/s)
^ I already got the answer for this one: 1.59×102 Hz
but please help me with part b
b)How long was the original pipe?
To calculate the length of the original pipe, we need to use the formula for the speed of sound in air:
v = f * λ
Where:
- v is the speed of sound in air (given as 345 m/s)
- f is the frequency of the lowest resonance (given as 244 Hz for one piece and 454 Hz for the other)
- λ is the wavelength
The formula for the wavelength of a sound wave in a pipe is:
λ = 2L/n
Where:
- L is the length of the pipe
- n is the mode or harmonic number (for the lowest resonance, n = 1)
We can rearrange the second formula to solve for the length of the pipe:
L = (n * λ) / 2
Now let's calculate the length of the pipe for each of the two resonance frequencies given:
For the first piece (frequency = 244 Hz):
λ = v / f = 345 / 244 = 1.415 m
L1 = (1 * 1.415) / 2 = 0.7075 m
For the second piece (frequency = 454 Hz):
λ = v / f = 345 / 454 = 0.7591 m
L2 = (1 * 0.7591) / 2 = 0.3796 m
To find the length of the original pipe, we need to add the lengths of the two pieces:
L_total = L1 + L2 = 0.7075 + 0.3796 = 1.0871 m
Therefore, the length of the original pipe is approximately 1.0871 meters.