Question

Name the following: Maximum value
of pressure of density of a sound
wave
Vour answer​

Answer 1

Answer:

Given -  ΔP=12sin(8.18x−2700t+π/4) ,

comparing this equation with , ΔP=Δp  

m

​  

sin(kx−ωt+ϕ) ,

we get ,  Δp  

m

​  

=12Pa ,

              ω=2700 ,

or            2πf=2700 ,

or            f=2700/2π

and          k=8.18 ,

or             2π/λ=8.18 ,

or             λ=2π/8.18 ,

therefore v=fλ=  

2π

2700

​  

.  

8.18

2π

​  

=330m/s

The relation between pressure amplitude Δp  

m

​  

 and displacement amplitude A (maximum value of displacement ) is given by ,

           Δp  

m

​  

=(vdω)A ,

where v= speed of sound in air ,

          d= density of medium (air) ,

          ω= angular frequency ,

given v=330m/s,d=1.29kg/m  

3

,ω=2700rad/s,Δp  

m

​  

=12Pa ,

now       Δp  

m

​  

=(vdω)A ,

or          A=  

vdω

Δp  

m

​  

 

​  

=  

330×1.29×2700

12

​  

=1.04×10  

−5

m

Explanation: