The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopr
anos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)
a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.
a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.
1 answer:
Answer:
Detailed step wise solution is attached below
Explanation:
(a) wavelength of the initial note 2.34 meters
(b) wavelength of the final note 0.389 meters
(d) pressure amplitude of the final note 0.09 Pa
(e) displacement amplitude of the initial note 4.78*10^(-7) meters
(f) displacement amplitude of the final note 3.95*10^(-8) meters
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Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:
where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,
<u>h = 593.50 m</u>
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:
where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,
Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:
where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,
<u>h₁₁ = 103 m</u>
c)
Now, we use first equation of motion for complete motion:
where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,
<u>vf = 107.91 m/s</u>