The speed of sound in air (m/s) depends on temperature according to aprox expression of

Question

v= 331.5 + 0.607 Tc

Tc= temp in celcius

In dry air, temperature decreases about 1deg celcius for every 150m rise in altitude.

a) assume the change is constant up to altitude of 9000 m.
What time interval is required for sound from airplane flying at 9,000m to reach ground on day when ground temperature is 30 deg celcius.

I tried to make a equation to do the problem and came up with

v= 331.5 + .607 (Tc- d/150)

and then I was thinking of plugging in
for

t= d/v after finding v

b) compare your answer with time interval required if air wre uniform at 30 deg celcius. Which is longer?

I think I would just plug into original equation 30 deg celcius

v= 331.5+ .607Tc
and then put v into t= d/v as well then compare them.

Is this alright?

Thanks

Answer 1

When flying at 9000 m, the sound speed at altitude is 331.5 - 0.607*(30-9000/150) = 331.5 - 18.2 = 331.5 m/s. On the ground the sound speed is 331.5+.607*30 = 349.7 m/s. The average speed of sound for a sound wave from plane to ground, when Tground - 30 C, is
(331.5 + 349.7)/2 = 340.6 m/s.
The answer is 9000 m/340.6 m/s= 26.42 s

For (b), forget about the decreasing T and sound speed with altitude. The sound speed all the way is
331.5 + 30*.607 = 349.7 m/s

The sound will get to the ground in less time in this case

Answer 2

I'm confused as to why the book says it takes 27.2 s for the wave to reach the ground for part a)

Thanks

Answer 3

And I'm also confused as to why you subtracted when the equation is addition?

v= 331.5 + .607 (Tc- d/150)

Thanks drwls

Answer 4

I wrote two negatives by mistake at one point. The temperature at flight altitude is -30 C, so the final speed of sound should be right. If I don't agree with the book's answer, I may have made another error in math somewhere. Sometimes the books are wrong.

Answer 5

Time T = integral of dt = Integral of dh/v from height h1 to height h2.

If v is a linear function of the height, you get the formula:

T = (h2 - h1)/(v2-v1) Ln(v1/v2)

In the case of this problem:

v1 = 349.71 m/s

v2 = 313.29 m/s

h1 = 0 m

h2 = 9000 m

So, T = 27.18 s

Answer 6

Count Iblis has found my mistake. I should have taken the average velocity to get time. Velocity appears in the denominator when you calculate elapsed time. In integrating to get elapsed time, you get a log term.

Answer 7

Thanks very much drwls and Count Ibis =D

Answer 8

I'm sorry. I know this question was asked a while ago, but I'm having trouble figuring out how you did your integration of dt
T = integral of dt = integral of dh/v from h1 to h2 (from 0 to 9000m)
did you use substitution dh = t dv?
Thanks!