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 1.0 kgm -3. Then pressure amplitude at r=√330 m from the point source is (using π=22 / 7) :

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 1.0 kgm -3. Then pressure amplitude at r=√330 m from the point source is (using π=22 / 7) : (A) 5 Nm -2 (B) 10 Nm -2 (C) 15 Nm -2 (D) 20 Nm -2

Tardigrade Admin · Accepted Answer

I=(P02/2 ρ V) ⇒ (P/4 π r2) =(P02/2 ρ V) ; where P, P0, V are power, pressure amplitude and velocity respectively. ⇒ P0=√(P ρ V/2 π r2) =√(50 π × 1 × 330/2 π × 330)=5

$5 N m^{- 2}$

$10 N m^{- 2}$

$15 N m^{- 2}$

$20 N m^{- 2}$

$I = \frac{P _{0}^{2}}{2 ρ V}$
$\Rightarrow \frac{P}{4 π r ^{2}}$
$= \frac{P _{0}^{2}}{2 ρ V};$ where $P, P_{0}, V$ are power,
pressure amplitude and velocity respectively.
$\Rightarrow P_{0} = \frac{Pρ V}{2 π r ^{2}}$
$= \frac{50 π \times 1 \times 330}{2 π \times 330} = 5$

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 1.0kgm−3. Then pressure amplitude at r=330​m from the point source is (using π=22/7) :

5Nm−2

10Nm−2

15Nm−2

20Nm−2

I=2ρVP02​​ ⇒4πr2P​ =2ρVP02​​; where P,P0​,V are power, pressure amplitude and velocity respectively. ⇒P0​=2πr2PρV​​ =2π×33050π×1×330​​=5

$5 N m^{- 2}$

$10 N m^{- 2}$

$15 N m^{- 2}$

$20 N m^{- 2}$

$I = \frac{P _{0}^{2}}{2 ρ V}$
$\Rightarrow \frac{P}{4 π r ^{2}}$
$= \frac{P _{0}^{2}}{2 ρ V};$ where $P, P_{0}, V$ are power,
pressure amplitude and velocity respectively.
$\Rightarrow P_{0} = \frac{Pρ V}{2 π r ^{2}}$
$= \frac{50 π \times 1 \times 330}{2 π \times 330} = 5$