If m=2000 p=2. 25 and y=6000 what is velocity
1 answer:
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The highest frequency sound to which the machine can be adjusted is :
- 4179.33 Hz
<u>Given data :</u>
Pressure = 10 Pa
Speed of sound = 344 m/s
Displacement altitude = 10⁻⁶ m
<h3>Determine the highest frequency sound ( f ) </h3>applying the formula below
Pmax = --- ( 1 )
Therefore :
f = ( Pmax * V ) /
= ( 10 * 344 ) / 2 * 1.31 * 10⁵ * 10⁻⁶
= 4179.33 Hz
Hence we can conclude that The highest frequency sound to which the machine can be adjusted is : 4179.33 Hz .
Learn more about Frequency : brainly.com/question/25650657
<u><em>Attached below is the missing part of the question </em></u>
<em>A loud factory machine produces sound having a displacement amplitude in air of 1.00 μm, but the frequency of this sound can be adjusted. In order to prevent ear damage to the workers, the maximum pressure amplitude of the sound waves is limited to 10.0 Pa. Under the conditions of this factory, the bulk modulus of air is 1.31×105 Pa. The speed of sound in air is 344 m/s. What is the highest-frequency sound to which this machine can be adjusted without exceeding the prescribed limit?</em>
Answer:
Explanation:
The amplitude of the oscillation under SHM will be .5 m and the equation of
SHM can be written as follows
x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.
x = .5 cosωt
given , when t = .2 s , x = .35 m
.35 = .5 cos ωt
ωt = .79
ω = .79 / .20
= 3.95 rad /s
period of oscillation
T = 2π / ω
= 2 x 3.14 / 3.95
= 1.6 s
b )
ω =
ω² = k / m
k = ω² x m
= 3.95² x .6
= 9.36 N/s
c )
v = ω
At t = .2 , x = .35
v = 3.95
= 3.95 x .357
= 1.41 m/ s
d )
Acceleration at x
a = ω² x
= 3.95 x .35
= 1.3825 m s⁻²