How Does Temperature Gradient Affect Sound Speed and Travel Time?

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SUMMARY

The discussion focuses on the relationship between temperature gradients and sound speed in air, specifically how a linear increase in temperature from T1 to T2 over a distance d affects sound wave travel time. The derived formula for sound speed as a function of distance incorporates temperature variations, expressed as v(general) = v[1 + (T1 + [(T2-T1)*x/d])/546]. The time taken for sound to travel through the temperature gradient is calculated using dt = dx/v, leading to a more complex expression involving the initial speed of sound at 273K and temperature adjustments.

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Q:The absolute temperature of air in a region linearly increases from T1 to T2 in a space of width d. Find time taken by sound wave to go through the region in terms of T1 T2,d and speed v of sound at 273K.

Equations :

vt = v( 1 + t/546)
dt = dx/v


Attempt:
I was basically looking to express speed of sound as a function of distance and integrating.

Assuming linear increase in temperature with distance
T(general) = T1 + [T2-T1]*x/d

Therefore
v(general) = v[ 1 + (T1 + [T2-T1]*x/d)/546 ]

dt = dx/v
= 546d (dx)/ v( 546d +T1d + {T2 -T1}x )

Am I right with this? I was hoping for an easier way as I am not sure if this is correct.
 
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I'm a little confused on what you are trying to do.

What is your v_t formula? Does t represent temperature for this formula? What is v?
 
Actually v is velocity and t is temperature so vt is velocity at that temperature.

vt/v273 = sqrt(T/273)
t=T+273 and using binomial expansion we have the result.
 

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