Calculating Train Whistle Frequency with a Moving Observer: Physics Problem Help

[buttterfly41]

i need help with waves :( please

the problem is:
While Jane waits on a railroad platform, she observes two trains approaching from the same direction at equal speeds of 8.60 m/s. Both trains are blowing their whistles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane, but before the second train passes her, she hears beats having a frequency of 4.60 Hz. What is the frequency of the trains' whistles? (Assume that the speed of sound in air is 343 m/s.)

ok, so this is as far as i can get:

F(beat)=4.6Hz V(train)=8.60m/s V=343m/s f' = trains whistle

f(beat)=f1-f2 f1=(f'*V)/(V+V(train) and f2=(f'*V)/(V-V(Train))

and putting it all togther i got:

f(beat)=[(f'*V)/(V+V(train)] - [(f'*V)/(V-V(Train))]
but this here doesn't work, but i don't know how else i would do this


any help would be much appreciated, thank you

 

[OlderDan]

the problem is:
While Jane waits on a railroad platform, she observes two trains approaching from the same direction at equal speeds of 8.60 m/s. Both trains are blowing their whistles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane, but before the second train passes her, she hears beats having a frequency of 4.60 Hz. What is the frequency of the trains' whistles? (Assume that the speed of sound in air is 343 m/s.)

ok, so this is as far as i can get:

F(beat)=4.6Hz
V(train)=8.60m/s
V=343m/s
f' = trains whistle

f(beat)=f1-f2
f1=(f'*V)/(V+V(train) and
f2=(f'*V)/(V-V(Train))

and putting it all togther i got:

f(beat)=[(f'*V)/(V+V(train)] - [(f'*V)/(V-V(Train))]
but this here doesn't work, but i don't know how else i would do this


any help would be much appreciated, thank you

The forum software does not preserve multiple spaces when you post, so I put your equations on separate lines in the quote.

Your approach is OK except that the right hand side of your last equation should be the absolute value of the difference. The way you have chosen to write the difference, it will be negative. If "doesn't work" means you are getting a negative result, just take the absolute value of the difference. If it is something else, check your computation and post your answer.

 

[buttterfly41]

The forum software does not preserve multiple spaces when you post, so I put your equations on separate lines in the quote.

Your approach is OK except that the right hand side of your last equation should be the absolute value of the difference. The way you have chosen to write the difference, it will be negative. If "doesn't work" means you are getting a negative result, just take the absolute value of the difference. If it is something else, check your computation and post your answer.

ok, so this is what i did,
F(beat)=4.6Hz
V(train)=8.60m/s
V=343m/s
f' = trains whistle

f(beat)=f1-f2
f1=(f'*V)/(V+V(train) and
f2=(f'*V)/(V-V(Train))

and putting it all togther i got:

f(beat)=[(f'*V)/(V+V(train)] - [(f'*V)/(V-V(Train))]
4.6=abs[343f'/(343+8.6)]-[343f'/343-8.6)]
4.6=abs[343f'/351.6]-[343f'/334.4]
4.6=abs[.9755f'-1.0251f']
4.6=.04956f'
f'=92.8Hz


ok, so this is the right answer...i was making calculator mistakes,

thanks again though :)