Question

A xylophone has a set of vibrating bars. A bar with a frequency of 343 Hz is struck. A very short time later, a second bar is also sounded, and one hears the volume of the xylophone increase and decrease 4 times each second (called beats). What are the possible frequencies of the second bar (in Hz)? (Give all correct answers, i.e. C, AE, ABCD, etc...) A) 335 B) 348 C) 347 D) 351 E) 339

Answer 1

I can't remember the equations here, but we want the peaks of the two waves to be aligned every quarter-second.

Answer 2

:S i am unsure as well, we havent leant this. hold on may this help

Answer 3

s(t)=[2scosw't]coswt

Answer 4

does that help? s1=scoswt s2=scoswt thats the resultant

Answer 5

Right, this equation comes from adding two waves together:
\[ \cos(f_1.t) + \cos(f_2.t) = 2 \cos(w_1.t).\cos(w_2.t) \]
where
\[ w_1 = \frac{f_1 + f_2}{2}, \ w_2 = \frac{f_1 - f_2}{2} \]

Now, if you're hearing beats, what you're hearing is the beats from the slower of these two frequencies, i..e, from \( w_2 \). Given that, you can now solve for the frequency of the other sound.

Answer 6

i still dont get it :/

Answer 7

Here's an example of beat behavior.
http://www.wolframalpha.com/input/?i=cos%28440t%29+%2B+cos%28445t%29

Answer 8

The point is, using superposition, if you have two sounds of a single frequency, then the sound you hear is just the sum of them: for example
cos(f1.t) + cos(f2.t)

Answer 9

That's the sound you would hear if you were hearing two sounds of equal amplitude, one of frequency f1, one of frequency f2. So far so good?

Answer 10

ya

Answer 11

Now, if f1 and f2 are close, then you will get "beats", as the two sounds interfere with each other. The beats can be calculated as I've shown above algebraically. The sound you hear as the beat sound, is the larger envelope of oscillation, the one with the lower frequency, the
\[ \cos\left( \frac{f_1 - f_2}{2}t \right) \] term. And in inside that larger oscillation is the othe other term, the cos[(f1 + f2)/2.t] term frantically oscillating up and down.

But

Answer 12

But what your ear is hearing as 'beats' is the longer oscillation, as the sound diminishes to zero, then peaks, then zero, then peaks, etc.

Answer 13

In your problem, your are hearing those oscillations 4 times a second. That means that
\[ \frac{f_1 - f_2}{2} \]
is such that you'll hear four beats. What must that number be equal to have four beats, or equivalently, four "zero" points?

Answer 14

i get what you mean, but you only have 1 frequency, i dont get the +/- 4/second

Answer 15

You have the original sound which has a frequency of f_1 = 343 Hz. Then a second sound is started with a frequency of f_2. Now you hear beats. So the question is: what values of f_2 make it possible for you to hear 4 beats/second?

Answer 16

So your ear is hearing cos(f1.t) + cos(f2.t)

Answer 17

wouldnt it b then ACDE?

Answer 18

What value must (f1 - f2)/2 have?

Answer 19

Since they will produce a difference of 4=|343-347|?

Answer 20

No. What you need is you need the function cos[(f1 - f2)/2.t] to have four beats, which is equivalent it having 4 zeros every second.

What value or values of (f1-f2)/2 make that possible?

Answer 21

OH its D and A no?

Answer 22

What value or values of (f1-f2)/2 ?

Answer 23

335 and 351

Answer 24

How many zeros does the function cos(2pi.t) have per second? 2.

Answer 25

How many zeros does the function cos(4pi.t) have per second? 4.

Answer 26

2

Answer 27

Hence you need
\[ \frac{f_1 - f_2}{2} = 4\pi \]

Now solve. And notice that cos is an even function so the change in frequency can be plus or minus.

Answer 28

so its 343-f2/2=4pi f2=343-8pi?

Answer 29

but that answer isnt even one of the opstions :/

Answer 30

And sorry: we need to normalize the original frequencies. If you have a sound wave with a frequency of f_1, that doesn't mean the function is cos(f1.t). It means the function is
\[ \cos(2\pi f_1 t) \]
So, we have
\[ \frac{ 2\pi |f_2 - f_1|}{2} = 4\pi \]

Answer 31

Therefore in fact,
\[ f_2 = f_1 \pm 4 \]

Answer 32

which yields 343-f2=4 so u get 347 and 339 correct?

Answer 33

Yes, \[ f_2 = 343 \pm 4 = 347, 339 \]

Answer 34

347 or 339

Answer 35

ok, i will look over the rules but i get what you mean

Answer 36

lol thx for taking so much time with me

Answer 37

It's worth playing with Wolfram to see what two sound waves of close frequency look like, and to convince yourself that the closer the sounds around, the less beats you hear per second.

Answer 38

If you ever play music or tune an instrument, this is very useful feature.

Answer 39

interesting. i took a look at the link u sent, i get what you mean. fact they had such a small difference they had fewer beats/s

Answer 40

Compare these two, with frequencies of 440 and 441:
http://www.wolframalpha.com/input/?i=cos%28440*2pi*t%29+%2B+cos%28441*2pi*t%29%2C+t+%3D+-1+to+1
vs. these two with frequencies 440 and 442:
http://www.wolframalpha.com/input/?i=cos%28440*2pi*t%29+%2B+cos%28442*2pi*t%29%2C+t+%3D+-1+to+1
vs. these two with frequencies 440 and 443:
http://www.wolframalpha.com/input/?i=cos%28440*2pi*t%29+%2B+cos%28443*2pi*t%29%2C+t+%3D+-1+to+1
and finally 440 and 450:
http://www.wolframalpha.com/input/?i=cos%28440*2pi*t%29+%2B+cos%28450*2pi*t%29%2C+t+%3D+-1+to+1