1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Significant figures vs precision (I think)

  1. Sep 3, 2015 #1
    1. The problem statement, all variables and given/known data
    I've got a doppler shift question about calculating the frequency heard by a receiver moving 5.0m/s towards a sound source that is emitting 3000Hz sound. I assumed speed of sound is 343 m/s for an air temperature of 20C.

    2. Relevant equations
    Freceiver = Fsender((V + Vreceiver)/V) Where V is speed of sound in air

    3. The attempt at a solution
    After calculating the sound I get 3000Hz x (348/343)
    Which equals 3000hz x 1.015
    If I round off to one or two significant digits like the problem states then I get 3000Hz. Which doesn't make sense. What am I missing here? Is there some rule the over rides significant digits for these types of answers?
  2. jcsd
  3. Sep 3, 2015 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    The number of significant figures in the given frequency of 3000 Hz is ambiguous. If you assume it has at least 3 significant figures, then you can keep 3 significant figures in the result of the calculation 3000 Hz x (348/343).

    If the 3000 Hz has two or less significant figures then you would get 3000 Hz for the answer. It just means that to two significant figures, there is no difference between the source frequency and the doppler shifted frequency.
  4. Sep 4, 2015 #3
    the point about significant figures, is that if you put data of limited accuracy into the equation; you get an answer of limited accuracy.

    To give a more accurate answer would be wrong (because, say, your 5.0m/s could mean 4.95m/s or 5.04 m/s - you don't know - all you can assume is that it was measured with an instrument of limited accuracy, so you cannot assume more accuracy).

    I would give the full calculation and write the answer as

    blah blah = 3043.731 Hz

    which is

    3.0 x 10^3 Hz to two significant figures

    3 x 10^3 Hz to one significant figure (assuming given frequency is correct to one sig fig)

    can't go wrong there - its not clear if the given frequency is correct to 1,2,3 or 4 figures ? so make an assumption, but make sure you write your assumption down.
    Last edited: Sep 4, 2015
  5. Sep 4, 2015 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Does the question ask for the frequency observed by the receiver, or does it ask for the frequency shift?

    Added in Edit:
    To further elaborate on my question:

    If you use significant digits (blindly) and use 2 significant digits, then the observed frequency gets rounded to 3000Hz.

    Calculating the Doppler shift from this may lead to the conclusion that the Doppler shift in frequency is 0 Hz. However, there is a way to calculate the shift without getting zero.

    ΔF = Freceiver - Fsender

    = Fsender((V + Vreceiver)/V) - Fsender

    = Fsender( ((V + Vreceiver)/V) - 1) ,​

    which can be further simplified..
    Last edited: Sep 5, 2015
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted