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 gwhereas the same volume of hydrogen has a mass of 2 g. Thus, the density of oxygen is 16 times thedensity of hydrogen at the same temperature and pressure. Asis same for both the gases,)oxygen()hydrogen(vv=)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. Thisenergycan be used to do work, for example, forcing the eardrums to vibrate or in the extreme case of a sonicboom 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 heldperpendicular to the direction of energy flow is called intensity.For a sound wave travelling along positive x-axis described by the equation.s = s0sin (t – kx +)P = p0cos (t – kx +)ts=s0cos (t – kx +)Instantaneous power P = F.v = pAtsP = p0cos (t – kx +) As0cos (t – kx +)Paverage= <P>= p0As0<cos2(t – kx +)>=2Asp00v =BB =v2p0= Bks0= =v2ks0
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PHYSICS"manishkumarphysics.in"6Paverage=kvpAp21200=v2Ap20=2spAv202maximum power = Pmax=vAp20= (pA) v2max,pv= pAv2s02Total energy transfer = Pav× t =2sAv202× tAverage intensity = average power / areathe average intensity at position x is given by<> =21vBs202=B2vP20....(5.1)Substituting B =v2, intensity can also be expressed as=v2P20....(5.2)Note :If the source is a point source thenI2r1and s0r1and s =rasin (t – kr +)If a sound source is a line source thenIr1and s0r1ands =rasin (t – kr +)Example 5.The pressure amplitude in a sound wave from a radio receiver is 2.0 × 10–3N/m2and the intensity at a pointis 10–6W/m2.If by turning the “Volume” knob the pressure amplitude is increased to 3 × 10–3N/m2, evaluatethe intensity.Solution :The intensity is proportional to the square of the pressure amplitude.Thus,=00p'por=00p'p=0.23× 10–6W/m2= 2.25× 10–6W/m2.Example 6.Acircular plate of area 0.4 cm2is kept at distance of 2m from source of powerW. Find the amount of energyreceived by plate in 5 secs.Solution :The energy emitted by the speaker in one second isJ. Let us consider a sphere of radius 2.0 m centeredat the speaker. The energyJ falls normally on the total surface of this sphere in one second. The energyfalling on the area 0.4 cm2of the microphone in one second=24–24104.0
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