Final answer:
The location of the mass, attached to the end of a spring oscillating with a period of 2.25s, at t=-5.515s is -5.515m. The mass is moving in the positive x direction at t=-5.515s with a velocity of 0.659m/s.
Step-by-step explanation:
To determine the location of the mass at -5.515 m, we can use the equation of motion for a mass-spring system. The general equation of motion is given by x = A * cos(ωt + φ), where x is the displacement, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase angle. In this case, the amplitude A is given as 0.0460 m and the period T is given as 2.25s. We can find the angular frequency ω using the formula ω = 2π/T.
ω = 2π / T = 2π / 2.25 = 2.79 rad/s.
Now we can find the phase angle φ using the initial condition x(0) = A * cos(φ).
0.0460 = 0.0460 * cos(φ).
Cancelling out the amplitude, we get cos(φ) = 1.
Since cos(0) = 1, the phase angle φ is 0.
Plugging in values into the equation of motion, we get x = 0.0460 * cos(2.79t).
To find the location of the mass at t = -5.515s, we substitute this value into the equation:
x = 0.0460 * cos(2.79 * -5.515) = -4.996 meters. Therefore, the location of the mass is -5.515 m.
To determine if the mass is moving in the positive or negative x direction at t = -5.515s, we need to find the velocity at that point in time. The velocity can be found by taking the derivative of the displacement equation with respect to time.
v = dx/dt = -0.0460 * 2.79 * sin(2.79 * -5.515) = 0.659 m/s.
Since the velocity is positive (0.659 m/s), the mass is moving in the positive x direction.