Answer:
68.8 Hz
137.6 Hz, 206.4 Hz
Explanation:
L = Length of tube = 2.5 m
v = Velocity of sound in air = 344 m/s
Distance between nodes is given by

Where n = 0, 1, 2, 3, ...
Making n+1 = n

where n = 1, 2, 3 .....
For fundamental frequency n = 1

Frequency is given by

The fundamental frequency is 68.8 Hz
First overtone

Second overtone

The overtones are 137.6 Hz, 206.4 Hz
Answer:
recall that heat absorbed released is given by
Q = mc*(T2 - T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-k)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed and (-) when heat is released.
substituting,
Q = (480)*(0.97)*(234 - 22)
Q = 98707 J = 98.7 kJ
Explanation:
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
