Respuesta :
Answer:
Part a)
Pressure amplitude is
[tex]P = 1.14 \times 10^[-2} Pa[/tex]
Part b)
displacement amplitude is given as
[tex]A = 7.49 \times 10^{-9} m[/tex]
Part c)
Distance at the other intensity is
[tex]r_2 = 63 m[/tex]
Explanation:
Part a)
As we know that intensity of sound at 5 m distance is 52 dB
so we have
[tex]52 = 10 Log(\frac{I}{I_o})[/tex]
now we have
[tex]5.2 = Log(\frac{I}{10^{-12}})[/tex]
[tex]I = 1.58 \times 10^{-7} W/m^2[/tex]
Now we know that speed of sound at 20 degree C is given as
[tex]v = 332 + 0.6 t[/tex]
[tex]v = 332 + 0.6(20)[/tex]
[tex]v = 344 m/s[/tex]
Now we know the relation of intensity of wave and pressure amplitude as
[tex]I = \frac{P^2}{2\rho v}[/tex]
[tex]1.58 \times 10^[-7} = \frac{P^2}{2(1.2)(344)}[/tex]
[tex]P = 1.14 \times 10^[-2} Pa[/tex]
Part b)
Now displacement amplitude with pressure amplitude is related as
[tex]P = BAk[/tex]
now bulk modulus of air is given as
[tex]B = 1.42 \times 10^5 Pa[/tex]
also we know that
[tex]\lambda = \frac{v}{f}[/tex]
[tex]\lambda = \frac{344}{587} = 0.586 m[/tex]
now we have
[tex]K = \frac{2\pi}{\lambda} = 10.72 m^{-1}[/tex]
so we have
[tex]A = \frac{1.14\times 10^{-2}}{(1.42 \times 10^5)(10.72)}[/tex]
[tex]A = 7.49 \times 10^{-9} m[/tex]
Part c)
As we know that intensity is inversely depends on the square of the distance
so we will have
[tex]L_1 - L_2 = 10 Log(\frac{I_1}{I_2})[/tex]
[tex]L_1 - L_2 = 10 Log(\frac{r_2^2}{r_1^2})[/tex]
[tex]52 - 30 = 10 Log(\frac{r_2^2}{5^2})[/tex]
[tex]r_2 = 63 m[/tex]
