- #1
- 18
- 3
- Homework Statement
- The 38 problem:The result on the book is 10.6cm i keep getting 3.75
- Relevant Equations
- X(t)=A×cos(w×t)
Last edited:
kuruman said:Please show your work so that we can figure out where you went astray.
kuruman said:Please show your work so that we can figure out where you went astray.
The only other equation i have is the one for the velocity in an instant of time v(t)=-vmax × sin(w × t).kuruman said:You are assuming that SHM is motion under zero acceleration. It is not. The velocity in SHM is not related to position and time by ##v=\frac{x}{t}##. What are the correct expressions for position and velocity in SHM? Your textbook should have those.
And yes, please learn how to write equations in LaTeX. Meanwhile, please do no post pictures sideways.
3.75*DaveC426913 said:How can you get a displacement of 37.5 if the max amplitude is specified as 15?
(This is intended to inspire a sanity check. If you mentally draw this motion in your head, that may help you figure out where you're going wrong.)
mm OK. Your notes say 3.75, though your problem statement says 37.5.Pawn said:3.75*
If your textbook really does not have an expression for the position as a function of time, look it up on the web.Pawn said:The only other equation i have is the one for the velocity in an instant of time v(t)=-vmax × sin(w × t).
W-omega
And i don't know what LaTeX is.
Do you have any idea on how to solve it?kuruman said:If your textbook really does not have an expression for the position as a function of time, look it up on the web.
To find how to use LaTeX to write and post equations, go here
https://www.physicsforums.com/help/latexhelp/
or click the link LaTeX Guide, lower left corner.
How ?hutchphd said:Knowing the period allows you to figure out ω. Knowing the position at t=0 allows you to figure out A. Then obtain x(t) for t=0.1s.
ω=2π/Thutchphd said:Look in your book for the relation between T and ω. Tell me what it is
But if we substitute the values we don't get the result that is in the book.hutchphd said:So in your equation for x $$x(t)=A\cos\frac {2\pi t} T $$ Now what is A?
Do we substitute pi with the value 3.14hutchphd said:cos (pi/4)= ?
No sorry i didnt realize ithutchphd said:You prefer 2pi/8 ? see #18
0.70hutchphd said:cos (pi/4)= ? ( that's 45deg)
Yes thank you so much for helping me .hutchphd said:So are you good? I got to go right now but let me know