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 forced into the open pipe,
L =
where n represents any whole number such as 1,2,3,4, etc., and λ denotes the wavelength of the oscillation
This implies λ=
It is important to note that n=1 corresponds to the fundamental frequency, n=2 corresponds to the first harmonic, and so forth.
Thus, the third harmonic will be for n=4
With L=6m and n=4, solving for λ yields:
λ= =3m
The connection between frequency (f), sound speed (c), and wavelength (λ) is given by:
c=f.λ or f=
≈115 Hz
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Answer:
Explanation:
Data provided
initial velocity v₀=20 cm/s at time t=3s
final velocity vf=0 at time t=8 s
Required
Average Acceleration for the interval from 3s to 8s
Solution
Acceleration can be defined as the first derivative of velocity concerning time
Answer:
Explanation:
Within a duration of 60 seconds, six waves are observed.
With a total of 6 waves,
this equates to 3 wavelengths.
As a result,
the period for each wavelength is calculated as 60 divided by 3.
Thus, period = 20 seconds.
According to the frequency-period relationship,
f = 1 / T
f = 1 / 20
f = 0.05 Hz
Response:
Clarification:
Refer to the diagram indicating the charges on the specified sphere (see attachment).
The electric field at the stated positions is
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We understand that electric potential correlates with the electric field through
V = Ed
A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Thus,
Vc = (kQ / c²) × c
Vc = kQ / c
As a result, the total charge Q consists of +q, -q, and +q
Hence, Q = q - q + q = q
V = kq / c
B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have
V = kQ/r
V = kQ / b, noting that r = b
So, Q = q
V = kq / b
C. At r = a
Following from equation 1:
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Thus,
V = Ed = 0 V
Consequently, the electric potential at the solid sphere's surface is 0.
D. At r = 0
The electric potential can be determined by
V = kq / r
As r approaches 0,
V = kq / 0
V approaches infinity.
