Question

the equation of a sound wave is y=0.0015sin(62.4x+316 t).find the wavelength of this wave​

Answer 1

Answer:

PHYSICS

The pressure variation in a sound wave in air is given by

ΔP=12sin(8.18x−2700t+π/4)N/m

2

Find the displacement amplitude. Density of air = 1.29kg/m

3

Give answer in terms of 10

−5

m

December 20, 2019avatar

Abhisri Waswani

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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

Answer 2

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