Convert Radians to Seconds - Solve f(t)=Rsin(ωt + θ)

[fonz]

Homework Statement

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

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

Homework Equations

N/A

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?

 

[Ray Vickson]

Homework Statement

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

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

Homework Equations

N/A

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?

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.

 

[LCKurtz]

Homework Statement

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

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

Homework Equations

N/A

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?

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.

 

[fonz]

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.

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.

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!

 

[Mark44]

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.

 

[fonz]

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.

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

 

[Mark44]

Yes.

 

[tiny-tim]

hey there, fonz!

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.

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) ​

 

[fonz]

Right sussed that then.

Thanks for all your help and contributions