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Radians to seconds?

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

    Given an equation such as f(t)=Rsin(ωt + θ)

    find the value of f(t) when t=5s

    2. Relevant equations

    N/A

    3. The attempt at a solution

    What I'm not understanding is that generally a function sin(ωt + θ) describes what happens when the angle t changes. However the question gives t in seconds.

    I'm pretty sure it is as simple as substituing 5 in for t but i'm not certain.

    My assumption is that knowing the frequency (ω/2∏ - I think!) allows you to know how many cycles occur in 1 second.

    So the frequency x t (5 x ω/2∏) gives you number of waves in 5 seconds - converting this to an angle by multiplying by 2∏ again leaves you with 5ω (ωt).

    Therefore seconds = radians?

    Can somebody confirm? Also - is my expression for frequency correct?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 20, 2013 #2

    Ray Vickson

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    Homework Helper

    You are mixing up angles and angular rates. The parameter ω has units of radians per second ant time t is in seconds, so ωt is in radians.
     
  4. Feb 20, 2013 #3

    LCKurtz

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    Gold Member

    The units on ##\omega## would be radians/second; it's the angular velocity. Multiply by t seconds and you get ##\omega t## in radians, add that to ##\theta## in radians and the units of the argument are radians. Then take the sine of it. :smile:
     
  5. Feb 20, 2013 #4
    Nooo - sorry I should have seen this coming...

    I chose ω arbitrarily to represent an angle. Yes I know ω is angular velocity usually but think of it as just an angle in radians (I should have picked a less confusing symbol)

    In any case it being the angular velocity wouldn't make any sense in what I had explained previously. I am simply asking if my assumption regarding the frequency and time is correct?

    I.e. is substituing 5 in for t going to give me the right answer!
     
  6. Feb 20, 2013 #5

    Mark44

    Staff: Mentor

    In your expression sin(ωt + θ), both θ and ωt are angles, typically in radians, so ωt + θ also represents an angle.

    Assuming you are measuring angles in radians, the only way that ωt can represent an angle when t is time is for ω to represent radians/time.
     
  7. Feb 20, 2013 #6
    I see what you are saying - this is a question for a course in electrical engineering. So presumably in the general case for sine wave signals the coefficient of t is always the angular velocity of the wave?

    So would the following statements be correct:

    For the function Rsin(40∏t + ∏)

    the frequency is 40∏/2∏ = 20Hz

    and the period would be 2∏/40∏ = 0.05
     
  8. Feb 20, 2013 #7

    Mark44

    Staff: Mentor

    Yes.
     
  9. Feb 20, 2013 #8

    tiny-tim

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    hey there, fonz! :cool:
    what's inside a sin (or cos or log or exp) must be an ordinary number

    (a radian is an ordinary number)

    sooo, if you have a time inside a sin (or log etc), it must be multiplied by a 1/time (or radian/time) :wink:
     
  10. Feb 20, 2013 #9
    Right sussed that then.

    Thanks for all your help and contributions
     
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