Calculating Ambulance Frequency Change: -113 Hz

  • Thread starter Thread starter Any Help
  • Start date Start date
  • Tags Tags
    Change Frequency
AI Thread Summary
An ambulance with a siren frequency of 880 Hz travels at 22 m/s, and the speed of sound is 343 m/s. The change in frequency heard as the ambulance passes is calculated using the Doppler effect formula. When the ambulance approaches, the observed frequency is 940 Hz, and when it moves away, it drops to 827 Hz. The difference in frequency is -113 Hz, confirming that the correct answer is D: -113 Hz. This illustrates how the Doppler effect alters sound frequency based on the relative motion of the source and observer.
Any Help
Messages
79
Reaction score
2

Homework Statement


:
An ambulance has a siren of frequency 880 Hz. You are sitting at a stop light as the ambulance passes you, traveling at a constant 22 m/s. If the speed of sound is 343 m/s, what is the change in the frequency that you hear as the ambulance passes you?


  • A

    -53 Hz
  • wrong-icon.png


    B

    -60 Hz


  • C

    -56 Hz
  • correct-icon.png


    D

    -113 Hz


  • E

    -116 Hz
the correct Answer is D : -113 Hz

Homework Equations


CodeCogsEqn-4.gif

WHERE FL and vL are the frequency and velocity of sound heard by the personL
fs and vs are the frequency and velocity of sound emmited by the source

The Attempt at a Solution


using the equation we get FL=880x(343+0)/(343-22) therefore FL=940.3Hz
the difference is FL-Fs=940.3-880=60.311Hz
but its not correct.
the correct answer is -113Hz why?
 
Physics news on Phys.org
What's the observed frequency when the ambulance is moving towards you? And then away from you?
 
Doc Al said:
WhaF frequency when the ambulance is moving towards you? And then away from you?
1) when towards me it is FL1=Fs*v/(v-vs)=880*343/(343-22)=940 Hz
when moves away from me FL2=Fs*v/(v+vs)=880*343/(343+22)=827Hz
FL2-FL1=827-940=113 Hz
Ah okay thanksssss
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top