Answer:
f = 4179.33 Hz
Explanation:
given,
Pressure = 10 Pa
bulk modulus of air = 1.31 × 10⁵ Pa
speed of sound = 344 m/s
displacement amplitude = 10⁻⁶ m
using formula
[tex]P_{max}=B(\dfrac{2 \pi f}{v})A[/tex]
[tex]f = \dfrac{P_{max}V}{2\pi BA}[/tex]
[tex]f = \dfrac{10\times 344}{2\pi \times 1.31 \times 10^5\times 10^{-6}}[/tex]
f = 4179.33 Hz
the highest frequency of the sound to machine is f = 4179.33 Hz
The highest-frequency sound to which the machine can be adjusted is :
Given data :
pressure = 10 pa
speed of sound = 344 m/s
displacement amplitude = 10⁻⁶ m
applying the formula below
Pmax = [tex]B(\frac{2\pi f}{v}) A[/tex]
Therefore ( f ) = Pmax * V / 2[tex]\pi \beta A[/tex]
= ( 10 * 344 ) / 2 *[tex]\pi[/tex] * 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
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