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) 1.301 kg/s
b) 0.001301 m³/s
c) V₁ = 6.505 m/s, V₂ = 1.626 m/s
d) 118.93 kPa
Explanation:
Given:
The number of cans = 220
The volume of can, V = 0.355 L = 0.355 × 10⁻³ m³
time = 1 minute = 60 seconds
gauge pressure at point 2, P₂ = 152 kPa
b) Thus, the volume flow rate, Q = Volume/ time
Q = (220 × 0.355 × 10⁻³)/60 = 0.001301 m³/s
a) mass flow rate = Volume flow rate × density
since it is mostly water, thus density of the drink = 1000 kg/m³
thus,
mass flow rate = 0.001301 m³/s × 1000 kg/m³ = 1.301 kg/s
c) Given:
Cross section at point 1 = 2.0 cm² = 2 × 10 ⁻⁴ m²
Cross section at point 2 = 8.0 cm² = 8 × 10 ⁻⁴ m²
also,
Q = Area × Velocity
thus, for point 1
0.001301 m³/s = 2 × 10 ⁻⁴ m² × velocity at point 1 (V₁)
or
V₁ = 6.505 m/s
for point 2
0.001301 m³/s = 8 × 10 ⁻⁴ m² × velocity at point 1 (V₂)
or
V₂ = 1.626 m/s
d) Applying the Bernoulli's theorem between the points 1 and 2 we have
or
on substituting the values in the above equation, we get
it is given that point 1 is above point 2 thus, y₂ -y₁ is negative
or
thus, gauge pressure at point 1 is 118.93 kPa