Doppler effect and Beat frequency

[Yoonique]

Homework Statement

A runner is running with speed V along the straight line connecting two identical speakers. Both speakers are playing a tone of the same frequency f.
a)What is the beat frequency that the runner hears?
b)If the standing wave forms between two speakers, how frequently will the runner hit a node of the standing wave?
c)Compare your results from (a) and (b).

Homework Equations

f_l = f_s(v+V_l)/(v+V_s), where v is the velocity of the wave, V_s is the velocity of the source and V_l is the velocity of the listener.
v = λƒ

The Attempt at a Solution

I solved part a using the doppler effect equation. T_beat = Vf/2v
For part b, I got the answer Vf/4v. Because it is two open ends, the start of the wave is an displacement antinode and thus the displacement node is at every λ/4.
But according to the answer it seems that the node they are referring to is the pressure nodes which are the displacement antinodes, thus the displacement antinodes is at every λ/2 which gives the answer Vf/2v which is the given answer.
So am I correct that the question refers the node as an pressure nodes? Or are they referring to the displacement nodes thus my concept is a bit shaky.

 

[paisiello2]

What exactly is a "pressure" node?

And why do you assume the wave is "open" at two ends? It actually doesn't matter.

Also using "T" for frequency is a bit confusing since this is typically used for period.

 

[haruspex]

the displacement node is at every λ/4.

Don't nodes and antinodes alternate? How can there be twice as many of one as of the other?

 

[Yoonique]

What exactly is a "pressure" node?

And why do you assume the wave is "open" at two ends? It actually doesn't matter.

Also using "T" for frequency is a bit confusing since this is typically used for period.

Okay shall change it to f_beat = Vf/2v. Because a wave equation is y(x.t) = Acos(kx-wt). And y represents the displacement of a particle from the equilibirum point. However in longitudinal wave, like sound wave, we can represent the wave in terms of a particle displacement or the pressure variations in the fluid where ΔP = BkAsin(kx-wt), where B is the bulk modulus of the fluid and k is the wave number (2π/λ). Then when the particle displacement from the equilibrium is 0 (displacement node), the pressure variation is the highest (pressure antinodes). That 2 equation have a phase difference of λ/4. So I'm asking that the question refers the node as a displacement node, or a pressure node. In a pressure node I assume there is not variation in pressure, thus you can't hear anything in a pressure node? While if you are in a displacement node, the variation in pressure is the largest thus the sound will be the loudest?

 

[Yoonique]

Don't nodes and antinodes alternate? How can there be twice as many of one as of the other?

Yeah they alternate. The wave formed between the 2 speakers is a standing wave with 2 open ends? So the displacement nodes are λ/4? Is kind of hard to explain so I'll use a picture.

 

[haruspex]

Yeah they alternate. The wave formed between the 2 speakers is a standing wave with 2 open ends? So the displacement nodes are λ/4? Is kind of hard to explain so I'll use a picture.

As per your picture, successive nodes are λ/2 apart, and successive antinodes are λ/2 apart.
The question asks how frequently, not how long to the first one.

 

[Yoonique]

As per your picture, successive nodes are λ/2 apart, and successive antinodes are λ/2 apart.
The question asks how frequently, not how long to the first one.

Okay I get what you mean. So that is why it does not matter whether is an open end because each successive nodes/antinodes are always λ/2 apart?

 

[haruspex]

Okay I get what you mean. So that is why it does not matter whether is an open end because each successive nodes/antinodes are always λ/2 apart?

Yes.