A point source of power 50 π watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m / s, atmospheric pressure is 1.0 × 105 Nm -2, density of air is (400/99 π) kgm -3. Then the displacement amplitude at r=√330 m from the point source is

Question

A point source of power 50 π watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m / s, atmospheric pressure is 1.0 × 105 Nm -2, density of air is (400/99 π) kgm -3. Then the displacement amplitude at r=√330 m from the point source is (A) 0.5 μ m (B) 0.2 μ m (C) 1 μ m (D) 2 μ m

Tardigrade Admin · Accepted Answer

P0=B K S0 ; k=(2 π/λ) ; λ=(v/f) ; v=√(B/ρ) Using above, we get S0=(P0/2 ρ v π f) =(5/2 × 1 × 330 × 3.14 × 1875) =1 μ meter

$0.5 μ m$

$0.2 μ m$

$1 μ m$

$2 μ m$

$P_{0} = B K S_{0}; k = \frac{2 π}{λ};$
$λ = \frac{v}{f}; v = \frac{B}{ρ}$
Using above, we get
$S_{0} = \frac{P _{0}}{2 ρ v π f}$
$= \frac{5}{2 \times 1 \times 330 \times 3.14 \times 1875}$
$= 1 μ$ meter

Q. A point source of power 50π watts is producing sound waves of frequency 1875Hz. The velocity of sound is 330m/s, atmospheric pressure is 1.0×105Nm−2, density of air is 99π400​kgm−3. Then the displacement amplitude at r=330​m from the point source is

0.5μm

0.2μm

1μm

2μm

P0​=BKS0​;k=λ2π​; λ=fv​;v=ρB​​ Using above, we get S0​=2ρvπfP0​​ =2×1×330×3.14×18755​ =1μ meter

$0.5 μ m$

$0.2 μ m$

$1 μ m$

$2 μ m$

$P_{0} = B K S_{0}; k = \frac{2 π}{λ};$
$λ = \frac{v}{f}; v = \frac{B}{ρ}$
Using above, we get
$S_{0} = \frac{P _{0}}{2 ρ v π f}$
$= \frac{5}{2 \times 1 \times 330 \times 3.14 \times 1875}$
$= 1 μ$ meter