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:
Explanation:
The path length difference = extra distance traveled
The destructive interference condition is:
where m =0,1, 2,3........
So, ←
⇒ λ = 2Δd = 2×10 = 20
Answer:
154° at 195 km/h
Explanation:
The helicopter is moving south at 175 km/h, relative to the wind.
But the wind is moving east at 85 km/h, relative to the ground.
This means that the helicopter is moving south east relative to the ground.
Every hour, the helicopter will move 175 km to the south and 85 km to the east, relative to the ground.
This means that we can determine the speed and direction of the helicopter using a right triangle and simple trigonometry.
Refer to the triangle b1.
The distance traveled by the helicopter in 1 hour is denoted by d.
d is the hypotenuse of the right triangle.
Using the Pythagorean Theorem, we can calculate d to be 195 km (rounded to 3 s. f.)
Hence the helicopter is traveling at 195 km/h relative to the ground.
To calculate the direction we use,
tan (x) = opposite/adjacent = 85/175
So the angle x is,
= 25.9°
Relative to the North, the helicopter is moving at 180° - 25.9° = 154° (rounded to 3 s. f.)
