Maximizing Speed in Simple Harmonic Motion: Solving for Displacement

[Soaring Crane]

An object attached to a spring (k = 30 N/m) has a velocity of 2.5 m/s when it is .55 m from equilibrium, and has a period of 2.45 s.

After doing other parts I found the mass to be 4.561 kg, the total energy of the system to be 18.79 J, the amplitude to be 1.12 m, and the maximum speed to be 2.86 m/s.

The part that I am stuck on/don't understand is: At what displacement is the speed maximized?

What am I supposed to do? What formulas am I supposed to use?

 

[dextercioby]

An object attached to a spring (k = 30 N/m) has a velocity of 2.5 m/s when it is .55 m from equilibrium, and has a period of 2.45 s.

After doing other parts I found the mass to be 4.561 kg, the total energy of the system to be 18.79 J, the amplitude to be 1.12 m, and the maximum speed to be 2.86 m/s.

The part that I am stuck on/don't understand is: At what displacement is the speed maximized?

What am I supposed to do? What formulas am I supposed to use?

I made use of a simpler version of it in the other thread.
It is:
v(t)=\omega A\cos(\omega t+\phi)
,from where u must find the 'cos'.From the 'cos',it's easy to find the 'sine' which is going to be needed to compute the "x".
x(t)=A\sin(\omega t+\phi)

Daniel.

 

[wais]

To solve for the displacement at which the speed is maximized, you can use the formula for maximum speed in simple harmonic motion, which is vmax = ωA, where ω is the angular frequency and A is the amplitude.

First, we need to find the angular frequency, ω, which can be calculated using the formula ω = 2π/T, where T is the period of the motion. In this case, T = 2.45 s, so ω = 2π/2.45 = 2.566 rad/s.

Next, we can use the given amplitude of 1.12 m to solve for the maximum speed, vmax = 2.566 rad/s * 1.12 m = 2.87 m/s.

Finally, we need to find the displacement at which this maximum speed occurs. To do this, we can use the formula for displacement in simple harmonic motion, x(t) = A*cos(ωt), where x(t) is the displacement at time t. We know that at t = 0, the object is at equilibrium (x = 0), so we can set up the equation as 0 = 1.12*cos(2.566*0). Solving for cos(0), we get x(t) = 1.12.

Therefore, the displacement at which the speed is maximized is 1.12 m. This makes sense intuitively, as the object's speed will be highest at the point where it is farthest from equilibrium, which in this case is at a displacement of 1.12 m.