Question
What is the length of the pipe?
Answer:
(a) 0.52m
(b) f2=640 Hz and f3=960 Hz
(c) 352.9 Hz
Explanation:
For an open pipe, the velocity is given by
Making L the subject then
Where f is the frequency, L is the length, n is harmonic number, v is velocity
Substituting 1 for n, 320 Hz for f and 331 m/s for v then
(b)
The next two harmonics is given by
f2=2fi
f3=3fi
f2=3*320=640 Hz
f3=3*320=960 Hz
Alternatively, and
(c)
When v=367 m/s then
Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
To solve this problem we will apply the concepts related to energy conservation, so the potential energy in the package must be equivalent to its kinetic energy. From there we will find the speed of the package in the vertical component. The horizontal component is given, as it is the same as the one the plane is traveling to. Vectorially we will end up finding its magnitude. So,
Here,
m = Mass
g = Gravity
h = Height
v = Velocity
Rearranging to find the velocity
Replacing,
Using the vector properties the magnitude of the velocity vector would be given by,
Therefore the package is moving to 66.2m/s