Solving Military HW: Position & Speed at Time 0s

[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.

 

[rwisz]

I would first clarify some information about the system being tested. Is the military HW undergoing simple harmonic motion (SHM) or is it a different type of motion? If it is SHM, then the equations and concepts of SHM can be applied to solve this problem.

To find the time at which the HW passes through 15cm, we can use the equation for displacement in SHM: x(t) = A*cos(omega*t), where A is the amplitude and omega is the angular frequency. Plugging in the given values, we get 15 = 23*cos(346*t). Solving for t, we get t = 0.0027 seconds. Therefore, the HW passes through 15cm at 0.0027 seconds.

To find the position when the speed is 200ms^-1, we can use the equation for velocity in SHM: v(t) = -omega*A*sin(omega*t). Plugging in the given values, we get 200 = -346*23*sin(346*t). Solving for t, we get t = 0.0027 seconds. Therefore, the position when the speed is 200ms^-1 is 23cm.

In SHM, the movement can be thought of as a point moving in a circle, but the motion is only in one dimension (along the x-axis in this case). The unit circle can be used to understand the trigonometric functions involved in SHM, but it is not necessary to solve this problem.

Overall, the key concepts to keep in mind when solving this problem are the equations for displacement and velocity in SHM and the relationships between amplitude, frequency, and angular frequency.