Doppler Effect Homework: Source Moving Away from Observer

[thornluke]

Homework Statement

A source of sound emits waves of wavelength λ, period T and speed v when at rest.
The source moves away from a stationary observer at speed V, relative to the observer. The wavelength of the sound waves, as measured by the observer is

A. λ + vT
B. λ - vT
C. λ + VT
D. λ - VT

Homework Equations

f' = f(v / (v±u) moving source
f' = f((v±u) / v) moving observer

The Attempt at a Solution

Since the source is moving away, it is safe to say assume that the frequency would be lower as the wavelengths emitted from the moving source would become more and more apart.

Therefore f' < f of moving source.

But up till here, I am lost...Please explain! I want to understand this question.

Cheers.

 

[superdave]

Well, if you know λ will be larger, that eliminates 2 answers. so the only question is whether λ will increase be vT or VT. Which makes more sense?

 

[thornluke]

Well, if you know λ will be larger, that eliminates 2 answers. so the only question is whether λ will increase be vT or VT. Which makes more sense?

C since v is the speed at rest, whilst V is the speed when the source is moving away.
Is it possible to explain this by deriving different equations to λ + VT?

 

[haruspex]

C since v is the speed at rest, whilst V is the speed when the source is moving away.

v is the speed of sound when the source is at rest, V is the speed of the source.
A question for you: what is the speed of the sound, as judged by the listener, when the source is moving away?

Is it possible to explain this by deriving different equations to λ + VT?

I don't understand the question. Are you asking how to derive the formula λ + VT?
Consider two successive wavefronts leaving the source. The first leaves the source. After time T it has advanced vT (=λ) towards the listener. Meanwhile, the source is VT further away. So when the second leaves the source it is VT+vT behind its predecessor. They will stay that far apart all the way to the listener.