Trig Waveform Question

[znowman]

Hi,

 

I've been looking at the following problem and wonder if anyone can help me out a little:

 

The following sinusoidal wave represents the fatigue stress in an axle   σ = 200sin(100π t + 0.25π)

a ) State the amplitude, angular velocity, periodic time, frequency and phase angle of the wave.

b ) How many seconds does it take for the stress to reach its max value?

 

These are the answers i've come up with for part a:

Amplitude = 200

Angular velocity = 100π

Periodic time = 2π / 100π = 1/50  (20 milliseconds)

Frequency = 1/1/50 = 50Hz

Phase angle = 0.25π / 100π = 1/400 radians (0.14°)

 

Part b is where i'm struggling, and I originally thought the first peak (maximum value) would be equal to periodic time/4 = 20 milliseconds/4 = 5 milliseconds

but then i've realised the wave is leading because of the +0.25π

 

I've tried 2π / 0.25π which gives me 8 (radians i assume), but i'm not sure where to go next... and wondered if anyone can help?

 

Thanks for looking, and any help or a pointer in the right direction would be much appreciated.

Edited January 2, 2018 by znowman

[sanctus]

Not looking into your a rplies, but for b:
isn't it just getting the values for t for which 100\pi t + 0.25 \pi =\pi/2 ?

I mean t starts from zero and 0.25\pi is were sine is growing, so solving for \pi/2 is the first maximum.

[znowman]

Hi, and thanks for getting back to me.

 

I've only just had chance to look at this, so havent digested what you are saying, but do the backslashes (\) in your post mean multiplication?  - i'm easily confused.

 

I did use CAD's 'curve by formula' tool to generate the attached graph:

 

 

It seems the answer is 2.5 milliseconds - just need to do the maths behind it now.

 

Thanks again for the reply, will have a look at what you said.

[sanctus]

Sorry that is from LaTEX notation (look it up if you do not know what it is, it is what most scientific articles are written in).

So \pi is just [math]\pi[/math]
And if I put the proper tags around my text (namely square [ math ] my stuff [ /math ] withouth the spaces) then it becomes:

[math]100\pi t + 0.25 \pi =\pi/2[/math]


Anyway your graph is right I get 1/400 sec as a result. Just did not want to give you the solution right away.

 

IF you get lost in this reasoning I posted let me know:
I mean t starts from zero and 0.25\pi is were sine is growing, so solving for \pi/2 is the first maximum.

[znowman]

Thanks, I'm not familiar with LaTEX but will have a look when i get chance.

 

so i guess if pi/2 is 1/2(pi) or 90 degrees, then 1/2(pi)/2 is 1/4(pi) or 45 degrees  which i think is the answer.  But i'm not sure how to get 1/4(pi) 0.785rads into seconds, it's the difference between angle and time that confuses me.

[sanctus]

For once, do not care about units ;).

At time t=0 the sine function is the growing phase (look at your plot, at 0.25 pi = 45 degrees --> sine is growing) so you want to find the time t for which your sine function reaches the first maximum. Sine reaches the maximum whenever its argument is of the form 0.5*pi + N* 2*pi with N any natural number. Since you are interested in the first maximum you can just set N=0 and hence you have to solve for t when the argument of your sine function equals 0.5*pi.
In steps:

 

[math]100\pi \cdot t +0.25\pi =0.5\pi [/math]

 

[math]100\pi \cdot t  =0.25\pi [/math]

[math]t =\frac{0.25\pi}{100\pi} [/math]

[math]t=1/400[/math]

 

[znowman]

I've been looking at your last post for a while, but i just can't make any sense of it in my own mind - as i said, i'm easily confused...

 

' Sine reaches the maximum whenever its argument is of the form 0.5*pi + N* 2*pi '

I don't understand where the 0.5*pi comes from, or where the N comes from?

 

sorry for being slow with this, i'm not very good with maths.

[sanctus]

No worries.

Ask yourself or your calculator when sine is 1, you will find that sin(pi/2)=1, but you will also find that that sine (5*pi/2)=1 and also sine (9* pi/2)=1 etc. and 5*pi/2= pi/2 + 1* 2*pi (so N=1) and 9*pi/2= pi/2 +2* 2*pi (N=2) . Note it would also work with negative values of N. And this is what means that sine has a period of 2*pi, you can add or remove as many multiples of 2*pi to the argument of sine and you get the same value.

TO complicate it a bit in question a) you are asked to find the period, it is not just 2*pi like I say just above because the argument is a function of t not just a value, but for any fixed value of t it is true. Say you put t=0 giving you sine (0.25 *pi) then what I say is still valid (sin(0.25*pi)==sin(0.25*pi+2*pi)) or for t=1 (sine(100*pi +0.25*pi)==sine(100*pi +0.25*pi+2*pi)) . Try it on your calculator.


So how do you find the period when the argument is a function? You do exactly the same approach, you know that sine has a period of 2 *pi so you look for time t' for which the difference is is 2*pi and since you want the time after t for which sine has the same value you can write t'=t+p, this leads to the following equation:
100*pi*(t+p)+0.25*pi -(100*pi*t +0.25*pi)=2*pi
[straight forward algebra, almost everything cancels out]
100*pi*p=2*pi
p=(2*pi)/(100*pi)

Which is what you did but now you might understand better why :-) I hope at least