Aliter the speed of sound in a gas is given by u p at

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Aliter : The speed of sound in a gas is given by u = P . At STP, 22.4 liters of oxygen has a mass of 32 g whereas the same volume of hydrogen has a mass of 2 g. Thus, the density of oxygen is 16 times the density of hydrogen at the same temperature and pressure. As is same for both the gases, ) oxygen ( ) hydrogen ( v v = ) hydrogen ( ) oxygen ( or, v (hydrogen) = 4v (oxygen) = 4 × 450 m/s = 1800 m/s. Ans. 5. INTENSITY OF SOUND WAVES : Like any other progressive wave, sound waves also carry energy from one point of space to the other. This energycan be used to do work, for example, forcing the eardrums to vibrate or in the extreme case of a sonic boom created by a supersonic jet, can even cause glass panes of windows to crack. The amount of energy carried per unit time by a wave is called its power and power per unit area held perpendicular to the direction of energy flow is called intensity. For a sound wave travelling along positive x-axis described by the equation. s = s 0 sin ( t – kx + ) P = p 0 cos ( t – kx + ) t s = s 0 cos ( t – kx + ) Instantaneous power P = F.v = pA t s P = p 0 cos ( t – kx + ) A s 0 cos ( t – kx + ) P average = <P> = p 0 A s 0 <cos 2 ( t – kx + )> = 2 A s p 0 0 v = B B = v 2 p 0 = Bks 0 = = v 2 ks 0
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PHYSICS "manishkumarphysics.in" 6 P average = k v p A p 2 1 2 0 0 = v 2 A p 2 0 = 2 s pAv 2 0 2 maximum power = P max = v A p 2 0 = (pA) v 2 max , p v = pAv 2 s 0 2 Total energy transfer = P av × t = 2 s Av 2 0 2 × t Average intensity = average power / area the average intensity at position x is given by <  > = 2 1 v B s 2 0 2 = B 2 v P 2 0 .... (5.1) Substituting B = v 2 , intensity can also be expressed as = v 2 P 2 0 .... (5.2) Note : If the source is a point source then I 2 r 1 and s 0  r 1 and s = r a sin ( t – kr + ) If a sound source is a line source then I r 1 and s 0 r 1 and s = r a sin ( t – kr + ) Example 5. The pressure amplitude in a sound wave from a radio receiver is 2.0 × 10 –3 N/m 2 and the intensity at a point is 10 –6 W/m 2 .If by turning the “Volume” knob the pressure amplitude is increased to 3 × 10 –3 N/m 2 , evaluate the intensity. Solution : The intensity is proportional to the square of the pressure amplitude. Thus, = 0 0 p ' p or  = 0 0 p ' p = 0 . 2 3 × 10 –6 W/m 2 = 2.25 × 10 –6 W/m 2 . Example 6. Acircular plate of area 0.4 cm 2 is kept at distance of 2m from source of power W. Find the amount of energy received by plate in 5 secs. Solution : The energy emitted by the speaker in one second is J. Let us consider a sphere of radius 2.0 m centered at the speaker. The energy  J falls normally on the total surface of this sphere in one second. The energy falling on the area 0.4 cm 2 of the microphone in one second = 2 4 2 4 10 4 . 0
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