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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Explanation:
given,
Angular speed of the tire = 32 rad/s
Displacement of the wheel = 3.5 rev
Δ θ = 3.5 x 2 π
= 7 π rad
now,
Time interval of the car to rotate 7π rad
using equation
Time taken to rotate 3.5 times is equal to 0.687 s.
Answer:
Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.
Explanation:
- The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
- As the car falls to the ground, its PE converts to KE.
- When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.
The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.
Assume that air resistance on the car is negligible. The height of the car at time
could be found with the equation:
,
where
near the surface of the earth, and
is the initial height of the car.
On the other hand, .
In other words, plotting the gravitational PE of the car against time would give a parabola. Since (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at
the initial GPE is positive, the
-intercept of this parabola should also be positive.
Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.
To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's . If there's no air resistance, the value of ME would stay at
Subtract PE from ME to obtain an expression for KE:
.
That's also a parabola when plotted against . Note that since the quadratic coefficient
is positive, the parabola shall open upwards.
