Frequency of varaiation of sound intensity

[somecelxis]

Homework Statement

A motorcyclist travel at cnstsant speed of 72km/h along a straight road toward an obesrver standing at the centre of bridge over the road . When the motor cyclist is distant, the obesrver hears a sound of 65hz from the engine of the motorcycle. What would be the frequency of sound heard by the obsrever after the motorcycle has passed under the bridge and away from obeserver? my ans is 57.5 hz. ( i managed to get the ans for this)
2nd part:
A second motorcyclist now rides alongside with part the first motorcyclist , at constsnt speed of 72km/h. The frequnecy f sound of 2nd motorsysle is lower than the first motorcycle. When 2 machines are treavelling towards the observer, from same direction , the intensity of sound varies with 3hz. What would be the frequency of variation of sound intensity after the motorcycles have passed the bridge and moving away from the obersver?

Homework Equations

The Attempt at a Solution

f ( motorclcye 2 heard ny observer) = 65-3 =62hz

to find frequency of mototrcycle 2 , i have 62/((330)/(330-20)) = 58.2 hz.

apparent frequncy of motorcycle 2 moving away from obsrever : ((330)/(330+20)) x 58.2 = 54.9 hz.

my ans is 57.5hz-54.9hz= 2.6hz.

but the ans is 3.7 hz . why i am wrong?

 

[somecelxis]

can anyone plese look at it and try to reply? thanks in advance!

 

[Simon Bridge]

Please show your reasoning.
The first part is about doppler shift, the second is more about beats.

 

[somecelxis]

Please show your reasoning.
The first part is about doppler shift, the second is more about beats.

i have shown my working and reasoning in the first post. can you help please?

 

[Simon Bridge]

You edited the post - OK.
Note: you got 57.5Hz for the first bike receding, and you know the second bike has a lower frequency than the first one... and yet you got a higher frequency (58.2Hz) for the second bike. Shouldn't the frequency still be lower when the bikes are receding?
What did you do in the first bike calculation that you did not do for the second bike?

Consider - what is the frequency of the second bike when it is stationary wrt the observer?

 

[somecelxis]

You edited the post - OK.
Note: you got 57.5Hz for the first bike receding, and you know the second bike has a lower frequency than the first one... and yet you got a higher frequency (58.2Hz) for the second bike. Shouldn't the frequency still be lower when the bikes are receding?
What did you do in the first bike calculation that you did not do for the second bike?

Consider - what is the frequency of the second bike when it is stationary wrt the observer?

this is actually what i mean . 65hz= sound heard by the obsrever when the bike 1 move towards him. the frequnecy of sound of bike 2 lower than bike 1 . so i assume the frequncy of sound of bike 2 heard by the obersver is lower . i have 65 hz-3hz = 62hz (sound frequency heard by the obbesrever when bike 2 move towards him. i also have the original frequnecy of the sound of bike 1 = 61.1hz which is higher than the bike 2 (58.2hz) , as shown in the working.

 

[Simon Bridge]

How did you calculate the correct frequency for bike 1 receding?

 

[somecelxis]

How did you calculate the correct frequency for bike 1 receding?

as shown in the photo , the frequncy when the bike 1 is receding is 57.5hz.
((330)/(330+20)) x 61.1 = 57.5hz.

 

[Simon Bridge]

as shown in the photo , the frequncy when the bike 1 is receding is 57.5hz.
((330)/(330+20)) x 61.1 = 57.5hz.

... where did you get each of those numbers from?
I'm guessing you are using 330m/s as the speed of sound. That would make the motorcycle speed 20m/s (=72kmph fine) ... so where did the 61.1Hz come from?

Now compare:

f ( motorclcye 2 heard ny observer) = 65-3 =62hz
to find frequency of mototrcycle 2 , i have 62/((330)/(330-20)) = 58.2 hz.

... This calculation is quite different from the one you did for motorcycle 1. Why did you do it differently?

I get the first step - ##f_1-f_2=f_{beat}## so you could get ##f_2## the frequency of motorcycle 2 while it is approaching. I'd like you to focus on that next step.

 

[somecelxis]

... where did you get each of those numbers from?
I'm guessing you are using 330m/s as the speed of sound. That would make the motorcycle speed 20m/s (=72kmph fine) ... so where did the 61.1Hz come from?

Now compare:... This calculation is quite different from the one you did for motorcycle 1. Why did you do it differently?

I get the first step - ##f_1-f_2=f_{beat}## so you could get ##f_2## the frequency of motorcycle 2 while it is approaching. I'd like you to focus on that next step.

i have found 61.1hz = frequency of original source bike 1 as shown in the photo

 

[Simon Bridge]

i have found 61.1hz = frequency of original source bike 1 as shown in the photo

OK - did you do that for bike 2?
i.e. what is the original frequency for bike 2?

You appear to be very reluctant to compare what you did for bike 1 with what you did for bike 2.
Why did you do a different calculation for bike 2 to what you did for bike 1?
This is what I want you to focus on.

You know - if you don't follow advise, I cannot help you.

 

[somecelxis]

... where did you get each of those numbers from?
I'm guessing you are using 330m/s as the speed of sound. That would make the motorcycle speed 20m/s (=72kmph fine) ... so where did the 61.1Hz come from?

Now compare:... This calculation is quite different from the one you did for motorcycle 1. Why did you do it differently?

I get the first step - ##f_1-f_2=f_{beat}## so you could get ##f_2## the frequency of motorcycle 2 while it is approaching. I'd like you to focus on that next step.

since##f_2## is the frequency of motorcycle 2 while it is approaching (not the original frequncy of sound bike 2 ) , then the ##f_1## in ##f_1-f_2=f_{beat}## means frequncy of bike 1 while it's approaching the observer . am i right? why do you say that my calsulation for part 2 is diffrent from part 1 ? in part 2 , i found the original frequncy of bike 2 from the the frequncy of sound heard when the bike 2 is approching the observer. i couldn't understand.

 

[somecelxis]

OK - did you do that for bike 2?
i.e. what is the original frequency for bike 2?

You appear to be very reluctant to compare what you did for bike 1 with what you did for bike 2.
Why did you do a different calculation for bike 2 to what you did for bike 1?
This is what I want you to focus on.

You know - if you don't follow advise, I cannot help you.

original frequncy for bike 2 = 58.2 hz .

 

[Simon Bridge]

Um - maybe I misunderstood... looks like I'll have to crunch your numbers.
Tomorrow...

Try this:
$$f_a=\frac{c}{c-v}f_0\\ f_r=\frac{c}{c+v}f_0$$ ... do you follow?
c is the speed of sound, v is the speed of the bike approaching or retreating for fa and fr respectively.

You are asked to find ##f_r## given ##f_a## with ##c## and ##v## ...
Best practice is to use simultaneous equations to eliminate the unknown ##f_0##.

Then plug in the numbers ...

 

[Simon Bridge]

Hmmm... I'm getting (310/350)x3=2.6571Hz ... perhaps the model answer is wrong after all.
Check my working. I may owe you an apology :)

[edit]
Just looking - the beats approaching is 3Hz, and they say the beats retreating is 3.7Hz ... that cannot be.
The frequency has to be lower right?
Note - my figure is about 2.7 .. maybe there's a typo turning 2.7 into 3.7?

Time to sleep.

 

[gneill]

Hi somecelxis,

Rather than keep you hanging until Simon has can get back to you, I can confirm that your result (~ 2.66 Hz) looks good. So the book's result is incorrect. Unfortunately this sort of thing happens more often than it should!

Note that you can take a shortcut to the result by multiplying the original beat frequency of 3 Hz by the two Doppler factors involved without having to obtain the individual frequencies of the bikes approaching and receding. Try it!

 

[somecelxis]

Hmmm... I'm getting (310/350)x3=2.6571Hz ... perhaps the model answer is wrong after all.
Check my working. I may owe you an apology :)

[edit]
Just looking - the beats approaching is 3Hz, and they say the beats retreating is 3.7Hz ... that cannot be.
The frequency has to be lower right?
Note - my figure is about 2.7 .. maybe there's a typo turning 2.7 into 3.7?

Time to sleep.

no aplogy needed. i really appreciate you effort to help. thanks

 

[Simon Bridge]

This is where my policy of not doing the numbers myself works against me.
When there is a possibility the model answer is wrong, that is where you need to get really pedantic about spelling out everything you do and why you chose to do it that way. Just in case the person setting the assignment made more than a typo.