Problem 5 7 of assignment value an orchestra contains both...

Problem 5: (7% of Assignment Value) An orchestra contains both string and wind instruments. The orchestra tunes their instruments in the morning at a temperature of TC = 15 \deg C. As they play, the temperature rises in the afternoon to TH 30 \deg C and falls in the late evening back to TC. As usual, the speed of sound in air increases with increasing temperature. Part (a) How has the increase in temperature from the morning to the afternoon altered the speed of sound in air?

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 Here it's given that in dry air the temperature decreases about 1 degree celsius for every 150 meter rise in altitude right now assume that this change is constant up to an altitude of 9 ,000 meters you have to calculate the time required for sound from airplane flying at 9 ,000 meters to reach to ground on a day when the ground temperature is 30 degrees celsius right great so first of all you have to write down the variation of the velocity of the sound with respect to temperature so there is a formula or expression we can say 331 .5 plus 0 .607 tc and is this is the temperature in celsius right now here we are given that the ground temperature is 30 degrees celsius right and every 150 degree rise in height there will be a decrease in 1 degree celsius right so this is a simple expression 30 minus y by 150 times of 1 degree right so this is the variation of temperature with altitude right why is the height now okay so we can write this expression i like this as a function of 5 right now simply we can plug in this value or at this velocity expression be equal to if you plug this tc then you will have 349 .71 minus 0 .0045 y right so this dc this value you have to just plug it in here and this expression of velocity with respect to the height right now we can directly write this expression velocity v equal to d y d t right and this is 349 point seven one minus 0 .00405 y right great now we have got d y over this expression which is 349 .71 minus 0 .00405 y and this is equal to dt right now you can take the integration on both the side this will be 0 to t we need to calculate the time rate and this wife will be from we have got 9000 to 0 because the sound will be from 9000 to 0 right height of 9 000 to 0 there we have the airplane now this comes out to be 1 by minus of this 1 .00405 natural logarithm of this expression which is 349 .71 minus 0 .00405 y right and this will be equal to t and this limit will be zero sorry 9000 here to zero now let's plug in and see how much we're getting this gives us t equal to uh 27 .2 seconds right so this is a time taken for the sound to move to the surface of the earth from 9 ,000 meters.

04:14 Next, there is a second part b...

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