What frequency will the bat hear?

[Lotto]

Homework Statement

A bat is flying with a speed $v$ perpendicular to a wall and is sending ultrasound waves of frequency $f_0$ and of speed $c$. The waves are reflecting from the wall. What frequency $f$ will the bat hear?

Relevant Equations

$f = f_0 \frac{2v}{c-v}$

The wavelength of the sent waves is $\lambda = \frac{c-v}{f_0}$. Reflected waves will maintain this wavelength. For the bat, the reflected wave's speed is $c+v$, so relative to the bat, its frequency is $\frac{c+v}{\lambda}=f_0\frac{c+v}{c-v}$. Since the difference between frequencies of the initial and the reflected waves is small, it will interfere and create beats of frequency $f=f_0\frac{c+v}{c-v}-f_0=f_0\frac{2v}{c-v}$.

In my textbook, the answer is $f=f_0\frac{v}{c-v}$, so I am confused. Is my solution wrong?

 

[phinds]

ultrasound waves of frequency $f_0$ and of speed $c$.

That's going to be one hell of a trick, getting ANY kind of sound wave, ultra or not, to travel at c. If you did not MEAN the speed of light, I suggest a different symbol.

 

[Lotto]

That's going to be one hell of a trick, getting ANY kind of sound wave, ultra or not, to travel at c. If you did not MEAN the speed of light, I suggest a different symbol.

Well, in the textbook, the symbol for the speed of sound in this particular task is $c$. I should have used a different symbol...

 

[phinds]

Well, in the textbook, the symbol for the speed of sound in this particular task is $c$. I should have used a different symbol...

The author should be shot.

 

[kuruman]

The author should be shot.

Should people who write things like $$\int x~dx=\frac{1}{2}x^2+c $$ also be shot?

 

[Steve4Physics]

It may be worth noting that (like it or not!) it is not uncommon for 'c' to be used for the speed of sound. This is typically seen where context prevents any confusion with the speed of light. E.g. https://en.wikipedia.org/wiki/Speed_of_sound

 

[WWGD]

Should people who write things like $$\int x~dx=\frac{1}{2}x^2+c $$ also be shot?

Ill leave that for those who use u,v , or i,j,k in a blackboard in the same lecture.

 

[hutchphd]

It just depends upon "whose ox is gored",,,,,,,,

 

[Lotto]

Guys, do you think that my solution is correct or is there a mistake?

 

[Orodruin]

That's going to be one hell of a trick, getting ANY kind of sound wave, ultra or not, to travel at c. If you did not MEAN the speed of light, I suggest a different symbol.

Using c for the speed of sound is extremely common. It is not that this author is peculiar here.

We could turn it around and suggest people using c to denote the constant 1 should be shot …

The given solution is obviously incorrect as the observed frequency would go to zero as $v\to 0$, which is absurd (it should obviously reproduce the emitted frequency).

 

[Steve4Physics]

....create beats of frequency $f=f_0\frac{c+v}{c-v}-f_0=f_0\frac{2v}{c-v}$.

In my textbook, the answer is $f=f_0\frac{v}{c-v}$, so I am confused. Is my solution wrong?

You are not asked ‘What is the beat frequency?’. You are asked ‘What frequency will the bat hear?’. IMO they are not (strictly speaking) the same.

Consider two sinusoidal signals, for simplicity $y_1(t) = \sin(2\pi f_1 t)$ and $y_2(t) = \sin(2\pi f_2 t)$. It is not difficult to show that:

$y_1 + y_2 = 2\cos(2\pi ~F_a~ t) \sin (2 \pi~ F_b~ t)$

where $F_a = \frac 12 |f_1-f_2|$ and $F_b = \frac 12 (f_1+f_2)$.

We are meant to assume that the bat hears the lower frequency ($F_a$) signal. This has frequency (using your Post #1 notation) of

$F_a=\frac 12(f_0\frac{c+v}{c-v}-f_0)=f_0\frac{v}{c-v}$

in agreement with the official answer.

The ‘beat frequency’ is in fact $2F_a$ corresponding to your answer. That’s where the confusion arose. You need to think about why this is so!

Minor edit.

 

[Barbarian]

Homework Statement: A bat is flying with a speed $v$ perpendicular to a wall and is sending ultrasound waves of frequency $f_0$ and of speed $c$. The waves are reflecting from the wall. What frequency $f$ will the bat hear?
Relevant Equations: $f = f_0 \frac{2v}{c-v}$

so relative to the bat, its frequency is c+vλ=f0c+vc−v.

When you observe moment when waves reflect from the wall you shouldn’t count it relative to bat. It wouldn’t matter because frequency of waves won’t change because of velocity of bat. So the new frequency of wave is c/f0 and so on as you already wrote. The main idea is not to solve this problem relative to bat. Its much more convenient to solve it relative to the wall

 

[Lotto]

You are not asked ‘What is the beat frequency?’. You are asked ‘What frequency will the bat hear?’. IMO they are not (strictly speaking) the same.

Consider two sinusoidal signals, for simplicity $y_1(t) = \sin(2\pi f_1 t)$ and $y_2(t) = \sin(2\pi f_2 t)$. It is not difficult to show that:

$y_1 + y_2 = 2\cos(2\pi ~F_a~ t) \sin (2 \pi~ F_b~ t)$

where $F_a = \frac 12 |f_1-f_2|$ and $F_b = \frac 12 (f_1+f_2)$.

We are meant to assume that the bat hears the lower frequency ($F_a$) signal. This has frequency (using your Post #1 notation) of

$F_a=\frac 12(f_0\frac{c+v}{c-v}-f_0)=f_0\frac{v}{c-v}$

in agreement with the official answer.

The ‘beat frequency’ is in fact $2F_a$ corresponding to your answer. That’s where the confusion arose. You need to think about why this is so!

Minor edit.

The waves sent by the bat and the waves reflected from the wall interfere. I thought that the net waves would be beats, which the bat then hears. I still don't understand why it hears a lower frequency...

 

[Steve4Physics]

The waves sent by the bat and the waves reflected from the wall interfere. I thought that the net waves would be beats, which the bat then hears. I still don't understand why it hears a lower frequency...

See if this helps. I’ve taken a diagram from http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/beat.html and added a few things.

The top part of the diagram shows two similar frequency signals (say $f_1$ and $f_2$).

The bottom part of the diagram shows the sum of these signals. The pattern is a low frequency signal (frequency $F_a = \frac 12 |f_1-f_2|$) modulating a high frequency ‘carrier’ (frequency $F_b = \frac 12 (f_1+f_2)$) as described in Post #11.

Left-to-right represents time. The periods of $F_a$ and $F_b$ are of course $T_a = \frac 1{F_a}$ and $T_b = \frac 1{F_b}$ which are marked on the diagram.

For humans (and presumably bats too!) the key signal that we would hear is the modulated signal, i.e. the signal of frequency $F_a$ enclosed in the modulation envelope (dotted lines).

Note that the lower diagram shows one full cycle - either dotted line is a complete sinusoid.
[EDIT: That's one full cycle of the modulation envelope.]

But what we experience during each period ($T_a$) of the modulating signal are two ‘beats’ (see diagram) corresponding to maximum amplitudes of the modulated carrier.

The time-interval between consecutive beats is $\frac {T_a}2$ so the beat frequency is $2F_a$.

I guess it boils down to how you choose to interpret the word 'hear'. Others may disagree with my interpretation!

 

[hutchphd]

I guess it boils down to how you choose to interpret the word 'hear'.

I believe the standard for most wave interactions is that the energy that is finally detected, and so classically one looks at the square of the amplitude, giving rise to the frequency doubling as you note. A common place it shows up is that fluorescent lights produce 120Hz interference from 60 Hz AC.

 

[kuruman]

This short article explains that bats modulate the signal frequency they emit to bring the Doppler-shifted echo within the sensitive range of their hearing. I didn't know they could do that!