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The equation of a sound wave is y=0.0015sin(62.4x+316 t).find the wavelength of this wave |
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Answer» PHYSICSThe pressure variation in a sound wave in air is given byΔP=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 mDecember 20, 2019avatarAbhisri WaswaniSHAREANSWERGiven - 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=2700/2πand k=8.18 ,or 2π/λ=8.18 ,or λ=2π/8.18 ,therefore v=fλ= 2π2700 . 8.182π =330m/sThe 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= vdωΔp m = 330×1.29×270012 =1.04×10 −5 m |
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