Velocity and height at each time interval

[mastermind1]

Homework Statement

I want to calculate velocity and height at each time interval
Time intervals: 1,2,3 (seconds)

equilibrium height = 3 meters
amplitude = 0.2
angular frequency = 48 rad/s
phase shift = 60

Homework Equations

y(t) = equiheight + Amplitude*sin(angfre*t + phaseshift)

The Attempt at a Solution

I have no idea about where to start. Please help.

 

[BvU]

Hello MM, and welcome to PF.
You could start with the height, for which you have presented a relevant equation. Nothing stops you from filling in equiheight, Amplitude, angfre, t and phaseshift, right ?
Then, for v you will need another equiation. No idea whatsoever ?

 

[mastermind1]

Hello MM, and welcome to PF.
You could start with the height, for which you have presented a relevant equation. Nothing stops you from filling in equiheight, Amplitude, angfre, t and phaseshift, right ?
Then, for v you will need another equiation. No idea whatsoever ?

Exactly. I know what you are saying about plugging everything in. What if we use Euler's algorithm to find the answer ? Any ideas about that ?

 

[BvU]

No can do: In the first place it isn't mentioned under 2. Homework Equations .
And if it were, why on Earth would it belong there ?

Do you know what to do to determine the velocity when an expression for the position is given ?
(Hint: consider the definition of velocity)

Euler's algorithm is for numerical integration.

[edit] And: if you "know what you are saying about plugging everything in", please show it, just to be sure things go alright...

 

[HalfLostAndSearching]

To calculate velocity and height at each time interval, we can use the equations of motion for simple harmonic motion. The equation for position as a function of time is given by y(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase shift. In this case, the equilibrium height is given by the constant term in the equation, so we can rewrite it as y(t) = 3 + 0.2*sin(48t + 60).

To find the velocity at each time interval, we can take the derivative of this equation with respect to time, which gives us v(t) = 0.2*48*cos(48t + 60). We can then plug in the values of t = 1, 2, and 3 seconds to get the velocity at each time interval.

For example, at t = 1 second, the velocity would be v(1) = 0.2*48*cos(48*1 + 60) = 9.6*cos(108) ≈ -3.9 m/s.

To find the height at each time interval, we can simply plug in the values of t = 1, 2, and 3 seconds into the original equation for y(t). For example, at t = 2 seconds, the height would be y(2) = 3 + 0.2*sin(48*2 + 60) ≈ 2.2 meters.

I hope this helps guide you in solving the problem. Remember to always identify the given variables and use the appropriate equations to find the desired quantities.