SHM with 23cm amplitude at 55Hz: when does it reach 15cm?

[jimmy42]

Homework Statement

A system is testing some military HW which has an amplitude of 23cm and frequency of 55Hz. At time 0s it is at the origin.

Homework Equations

At what time does it pass through 15cm?
When the speed is 200ms^-1 what is the position?

The Attempt at a Solution

omega = (2pi) * 55 = 346 r^s-1
max V = 23 * 346 = 7958 -- is that right??

To find the time, do I need to use trigonometry and the unit circle? No clues about the second one. Can SHM problems be found with the unit circle? Do I need to only use the SHM equations?

Should I think SHM in terms of the unit circle or movement going only up and down and not in a circle at all?

I'm really stuck, thanks for any help.

 

[rock.freak667]

You have that x=Xsin(ωt+φ) and you found that ω=346 rad/s and φ=0, X=23.

Hence you have x = 23sin(346t).

So when x=15, you just need to use standard trigs.

siny=a ⇒ y=sin^-1(a)


For the second one, velocity 'v' is related to 'x' by v = dx/dt.

 

[jimmy42]

Sorry, I don't get it.

What is a? I looked at that first equation but how do I find t from that?

Thanks,

Jim.

 

[rock.freak667]

Sorry, I don't get it.

What is a? I looked at that first equation but how do I find t from that?

Thanks,

Jim.

x = 23sin(346t).

then you have 15=23sin(346t). Make sin(346t) the subject and then take the inverse sine of that.