Problem with basic doppler effect question

[PsychonautQQ]

Homework Statement

The source of a 1 kilohertz sound is getting closer to the listener at a speed of .9 times the speed of sound. What frequency does the listener hear?

Homework Equations

fl = ((v-vs) / (v + vl))fs

The Attempt at a Solution

((343+343(.9)) / 343)*1000=1900 Hz
this is the wrong answer, what did I do wrong? it's supposed to be 10,000 Hz

 

[besulzbach]

Correct formula:
$f = \frac{c\ +\ v_{r}}{c\ +\ v_{s}} \cdot f_{0}$

In your problem the source is getting closer, so $v_{s}$ is negative.

As you didn't miss that, your error is in the formula. Use the one above.

You also SUMMED values that you should have SUBTRACTED.

Remember:

$v_{r}$ is the velocity of the receiver relative to the medium; positive if the receiver is moving towards the source;

$v_{s}$ is the velocity of the source relative to the medium; positive if the source is moving away from the receiver;

 

[STEMucator]

Homework Statement

The source of a 1 kilohertz sound is getting closer to the listener at a speed of .9 times the speed of sound. What frequency does the listener hear?

Homework Equations

fl = ((v-vs) / (v + vl))fs

The Attempt at a Solution

((343+343(.9)) / 343)*1000=1900 Hz
this is the wrong answer, what did I do wrong? it's supposed to be 10,000 Hz

Use this equation : $f = f_0( \frac{v}{v ± v_s} )$

Where $f$ is the stationary frequency. $f_2$ is the apparent frequency detected. $v$ is the speed of sound in air and $v_s$ is the speed of the source.

Since the source is moving closer to your listener, you want to use $v - v_s$ ( Doppler effect ). Don't forget to convert kHz to Hz otherwise this won't work!

I got $f = 10000 Hz$.