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I made up a question to make you understand my doubt , see

Assume that a car is travelling with speed v to a wall . When car is at distance x from wall it emits a sound signal . Find time after the signal reaches the car again ? ( Speed of sound $ = v_s$ )

Now if we think that sound wave is like an object and use relative motion than sound will approach wall with speed $v+v_s$ and again come to car after collision with wall with speed $v+2v_s$ So $$t=\frac{x}{v+v_s}+\frac{x-\frac{xv}{v+v_s} }{v+2v_s}$$

Is it expression right ? Also how will Doppler effect will affect the answer ?

Thanks for help and welcome for any edits!

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Now if we think that sound wave is like an object and use relative motion than sound will approach wall with speed $v + v_{s}$...

No, I think you are misunderstanding something here. A linear sound wave will always propagate at the speed of sound once emitted in a homogeneous, uniform medium. If your expression were correct, how could a shock wave form? That is, the assumption that the sound wave moves at $v + v_{s}$ implies that it will always be faster than the emitting object, which is not true.

...and again come to car after collision with wall with speed $v + 2 \ v_{s}$...

This second expression is also incorrect. The sound wave would approach the car at $v + v_{s}$.

When car is at distance x from wall it emits a sound signal. Find time after the signal reaches the car again ?

You need to solve this in a piece-wise fashion. Start with the following:

  • sound pulse emitted at $t = 0$ at a distance $\Delta x_{i}$ from the wall
    • this pulse will reach the wall in $\Delta t_{0} = v_{s} \ \Delta x_{i}$
  • at the time when this pulse hits the wall the emitting source has moved $\Delta x_{d} = v \ \Delta t_{0}$
    • at this time, the emitting source is $\Delta x_{r}$ from the wall
  • the final piece of time is given by $\Delta t_{1} = \Delta x_{r} \left( v + v_{s} \right)$

Then the answer to your question is just adding the two $\Delta t$'s together.

Also how will Doppler effect will affect the answer ?

The Doppler effect only changes the frequency at the receiver, not the speed of propagation.

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