# Relative velocity in the air frame or platform frame?

In "Introduction to Classical Mechanics: With Problems and Solutions" by David Morin he describes the following scenario:

Let $$v_s$$ be the speed of sound in air. Imagine two people standing on the ends of a long platform of length L that moves at speed $$v_p$$ with respect to the reference frame in which the air is at rest.

One person claps, the other person claps immediately when he hears the first clap (assume that the reaction time is negligible), and then the first person records the total time elapsed when she hears the second clap. What is this total time?

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Consider first the case where the platform moves parallel to its length. In the frame of the air, assume that the person at the rear is the one who claps first. Then it takes a time of $$\frac{L}{(v_s - v_p)}$$ for the sound to reach the front person. This is true because the sound must close the initial gap of L at a relative speed of $$v_s − v_p$$, as viewed in the air frame.

I have trouble understanding the part highlighted in bold. If I understood the scenario correctly, the air frame is where the velocity of the sound is $$v_s$$.

This mention of relative speed would make sense from the perspective of the observer moving in the platform. In that frame, the sound moves at $$v_s − v_p$$, the time it takes to move the distance L from his perspective is then $$\frac{L}{(v_s - v_p)}$$ as stated.

Did he mean to say "in the platform frame" instead of "in the air frame"?

Given the text the initial sound pulse moves in the same direction is the platform.

When using velocities I find that specifying what the velocity is measured relative makes things clearer.

So the velocity of the sound relative to the air is $$\vec v_{\rm sound,air}$$, the velocity of the person on the platform relative to the air is $$\vec v_{\rm person,air}$$, and velocity of the sound relative to the person on the platform is $$\vec v_{\rm sound,person}$$.

Thus $$\vec v_{\rm sound,\color{red}{person}} + \vec v_{\rm \color{red}{person},air}= \vec v_{\rm sound,air} \Rightarrow \vec v_{\rm sound,person} = \vec v_{\rm sound,air}- \vec v_{\rm person,air}$$

If it straight line motion $$v_{\rm sound,person} = v_{\rm sound,air}- v_{\rm person,air}$$ and $$v_{\rm sound,person}$$ is the speed of sound as measured on the platform of length $$L$$.

Thus the time, $$T$$, for the sound to traverse the platform is $$T = L/(v_{\rm sound,air}- v_{\rm person,air})$$.
you would expect the time to be longer than $$L/(v_{\rm sound,air}$$ as the person receiving the sound is moving away from the source of sound this making the sound travel further.

In this non-relativistic example that time, $$T$$, is the same for all observers so the reference frame does not matter.

because the sound must close the initial gap of L at a relative speed of $$v_{\rm s}-v_{\rm p}$$, as viewed in the air frame.

The gap measure by the person on the platform is always $$L$$ but the gap measured by a person standing on the ground (air at rest frame) is larger than $$L$$ because the front of the platform is moving in the same direction as the sound, ie making the distance the sound has to travel larger.

Another way of analysing the situation is to look at it from the ground (air at rest) frame.
It it takes a time $$T$$ for the sound go from the source to the observer at the other end of the platform the total distance traveled by the sound is $$L+ v_{\rm person,air}\,T$$ and this distance is also equal to $$v_{\rm sound,air}\,T$$.

$$L+ v_{\rm person,air}\,T=v_{\rm sound,air}\,T \Rightarrow T = L/(v_{\rm sound,air}- v_{\rm person,air})$$ as before.

with respect to the reference frame in which the air is at rest.

When we say "in X frame", X is definitely at rest in that frame, even if we have to resort to mathematical trickery define such a frame.

Air frame means the frame in which the air is at rest.