An organ pipe open at both ends has a radius of 4.0 cm and a length of 6.0 m. what is the frequency (in hz) of the third harmoni
1 answer:
When air is blown into the open pipe,
L =
where nis any integral number 1,2,3,4 etc. and λ is the wavelength of the oscillation
⇒λ=
Note here that n=1 is for fundamental, n=2 is first harmonic and so on..
⇒ third harmonic will be n=4
Given L=6m, n=4, solving for λ we get:
λ= =3m
Relationship of frequency(f), velocity of sound (c) and wavelength(λ) is:
c=f.λ Or f=
⇒f=
≈115 Hz
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Explanation:
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where x is the distance traveled with velocity v
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The distance traveled by grasshopper is
Answer:
a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm
b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm
Explanation:
Given data:
L = 20 mm
F = 250 N
r₁ = 10 mm
r₂ = 15 mm
v = 0.3
E = 2.07x10⁵ MPa
a) The maximum contact pressure is:
The width of contact is:
b) According the graph elastic stresses below the surface, for v = 0.3, the maximum shear stress is
T = 0.3*P = 0.3 * 274.58 = 82.37 MPa
At a distance of
0.8*b = 0.8*0.029 = 0.023 mm