An air-filled pipe is found to have successive harmonics at 945 Hz , 1215 Hz , and 1485 Hz . It is unknown whether harmonics bel
ow 945 Hz and above 1485 Hz exist in the pipe. What is the length of the pipe
1 answer:
Answer:
L = 0.635m
Explanation:
This problem involves the concept of stationary waves in pipes. For pipes closed at one end,
The frequency f = nv/4L for n = 1,3,5....n
For pipes open at both ends
f = nv/2L for n = 1,2,3,4...n
Assuming the pipe is closed at one end and that velocity of sound is 343m/s in air. If we are right we will obtain a whole number for n.
The film solution can be found in the attachment below.
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Answer:
Δy=0.431m
Explanation:
Diffraction grating with split space d,to find the fringe position ym,we must to find the angle from
dSinα=mλ
A grating with N slits or lines per mm has slit spacing of
d=1/N
d=(1/600mm)
d=1.67×10⁻³mm
For 400nm wavelength:
α=Sin⁻¹(mλ/d)
And the position of first order lowest wavelength fringe on the screen is:
For 700nm wavelength:
α=Sin⁻¹(mλ/d)
And the position of first order highest wavelength fringe on the screen is:
The difference between the first order lowest and highest wavelength fringe is
Δy=(0.925595 - 0.49445)m
Δy=0.431m