Question

A whistle of frequency f0 = 1120 Hz moves in a circle of radius R = 2.00 m at an angular velocity of omega = 31.5 rad/s.
a) What is the maximum frequency heard by a stationary listener in the plane of the circle and 6.20 m away from its center?
b) What is the minimum frequency heard by the listener?

Answer 1

this question it related to doppler's effect

Answer 2

Ok, so first of all if the source is emitting sound of frequency \( f_0 \), write as f0, what is the formula for the frequency heard by an observer, where
v = speed of sound
vs = velocity of source
vl = velocity of listener

f = ... what?

Answer 3

still there?

Answer 4

idk

Answer 5

yes

Answer 6

f=(v-v0/v+vs)*f0

Answer 7

(340-v0)/(340+vs)*1120

Answer 8

right. Now, the listener isn't moving so you zero that out. All you need now is an expression for the velocity of the source.

Answer 9

ok so it would be: (340/340+vs)*1120. but we have two unknowns? :S how can we find vs?

Answer 10

Remember that equation of motion in a circle about the origin can be written as

JamesJ: r(t) = A(cos(omega.t), sin(omega.t))

JamesJ: Radius of motion is A

JamesJ: period is 2.pi/omega. And hence frequency is omega/2.pi

Answer 11

Suppose therefore the listener is sitting at (6.2,0).
The position of the whistle is
\[ s(t) = 2(\cos(31.5t), \sin(31.5t)) \]
and hence the velocity of the whistle is ...

Answer 12

Now. The only motion the observer cares about is motion away from him. Now the distance of the whistle from the observer is
\[ d(t) = \sqrt{ (6.2 - 2 \cos(31.5t))^2 + (2\sin(31.5t)^2 } \]
Simplify this expression a little and find find the speed of the whistle relative to the observer. Make sure the speed is +ve if the whistle is moving towards the listener and -ve if the whistle is moving away.

Answer 13

this expression above will give me the velocity of the source? for maximum we would subtract and minimum add.

Answer 14

That is the distance, using Pythagorus theorem, or just good old fashioned Euclidean distance.

It can be simplified. Now differentiate it to find the speed.

Answer 15

Here's a diagram. Make sense?

Answer 16

i will try an answer, if it does not work, then i will let you know. :P

Answer 17

since we will have found our distance sing the distance formula, we plug it back into s(t) and then from there we plug it back into our formula which should give us a maximum by subtracting and minimum by adding correct?

Answer 18

No, you're making it too complicated.

Answer 19

We need to know the speed of the whistle relative to the listener. That is the time derivative of the distance between them d(t).

That derivative \( \frac{d \ }{dt} d(t) \), is the velocity \( v_{source} = v_{whistle} \)

Answer 20

Notice that
\[ d(t) = [ 6.5^2 + 2^2 (\cos^2(31.5t) + \sin^2(31.5t)) - 2\cdot 2 \cos(31.5t) ]^{1/2} \]
\[ = [ 46.25 + 4 \cos(31.5t) ]^{1/2} \]

Answer 21

Hence
\[ v = d'(t) = \frac{1}{2}\frac{-126 \sin(31.5t)}{[46.25+4\cos(31.5t)]^{1/2}} \]

Answer 22

Now the velocity to the observer is
\[ f = \frac{v_{sound}}{v_{sound} + v}f_0 \]
where v is the equation above.

It's obvious that frequency f will be maximized when v is minimized and vice versa.

Answer 23

So find the max and min of the function v = v(t), as above; those max and min will correspond respectively with the min and man of the frequency f.

Answer 24

but what is t?

Answer 25

t is time.

Answer 26

i know, but you do not have it

Answer 27

You know calculus, yes?

Answer 28

yes

Answer 29

So you need treat v(t) as a function in t and find the local max and min. by differentiating, setting it equal to zero, solving for the t when the derivative is zero, etc.

Or, if you're good at Physics, you can figure it out without calculus.

But for now, you need to bolster your intuition and follow the equations. So differentiate away.

Answer 30

is there not a simpler way of doing this problem? this whole approach seems rather much more complicated.

Answer 31

since i have to find local max and min, i need to take the double derivative of v?

Answer 32

You want to find the maximum and minimum values of v(t).

here's a way to do it without calculus.

As far as the listener is concerned, v(t) will be maximized when the whistle is moving exactly towards him. In other words, the velocity vector of the while lies along the tangent line of the circle that intersects with the listener.

Likewise, the while is moving away the highest speed when the whistle has the instantaneous velocity which is in the direction of the other tangent line which intersects with the listener; but moving away from the listener.

In both cases, the speed of the whistle is
\[ v = \omega r \]
where \( \omega \) is the angular velocity, 31.5

Answer 33

Are you in college btw?

Answer 34

yes 2nd semester

Answer 35

right

Answer 36

ya that is what i was thinking, but i am just uncertain about the r. and btw are u in university?

Answer 37

I graduated a while ago. Mythias, read the question ... you are told the value of r !

Answer 38

"A whistle of frequency f0 = 1120 Hz moves in a circle of radius R = 2.00 m"

Answer 39

ya but where does the 6.2m play a role in?

Answer 40

We used that when we wrote down the distance function d(t) and it now appears de facto in the velocity function v(t)

Answer 41

if we do v=wr. we get the velocity of the whistle. couldnt we just plu in to the doppler formula with that information? i am still confused with regards to the 6.2m.

Answer 42

NVM I GET IT LOL

Answer 43

james, i will post some random things up allow me to give u some medals. i owe u a lot especially for all the time that you spent with me. i really appreactiate