The position of a 0.31 kg object attached to a spring is described by [x = (0.85 m) sin(2.37πt + 0.35)]
a) What is the amplitude of oscillations in m?
b) What is in meters/second the maximum speed of the oscillating object? Answer in meters/seconds .
c) Find in meters/second the object’s speed at t = 0.98 s.
d) What is in meters/seconds2 the maximum acceleration of the oscillating object?
I will be happy to critique your thinking on this.
To answer these questions, we will use the given equation and some basic principles of harmonic motion. Let's break it down step by step:
a) The amplitude of oscillations can be determined from the given equation. In this case, the amplitude is the coefficient in front of the sine function. From the equation, the amplitude is 0.85 m.
b) The maximum speed of an oscillating object occurs when the displacement from the equilibrium position is maximum. This means we need to find the derivative of the position equation with respect to time, and then evaluate it at the maximum displacement.
Taking the derivative of the given equation, we get:
v = dx/dt = (0.85 m) * (2.37π) * cos(2.37πt + 0.35)
To find the maximum speed, we need to evaluate this equation when the object is at the maximum displacement. From the equation, we know that the maximum displacement is 0.85 m. Plugging this value into the equation, we get:
v = (0.85 m) * (2.37π) * cos(2.37πt + 0.35)
When t = 0, the object is at the maximum displacement, so we plug in t = 0:
v = (0.85 m) * (2.37π) * cos(0.35)
Calculating this expression will give us the maximum speed of the oscillating object in meters/second.
c) To find the object's speed at a specific time t = 0.98 s, we need to plug this value into the velocity equation we derived earlier:
v = (0.85 m) * (2.37π) * cos(2.37π * 0.98 + 0.35)
Evaluating this expression will give us the object's speed at t = 0.98 s.
d) The maximum acceleration of the oscillating object can be found by taking the derivative of the velocity equation with respect to time:
a = dv/dt = -(0.85 m) * (2.37π)^2 * sin(2.37πt + 0.35)
To find the maximum acceleration, we need to evaluate this equation at the maximum displacement, which is 0.85 m. Plugging in this value and evaluating the expression will give us the maximum acceleration of the oscillating object in meters/second^2.
Remember to use the proper units and carry out the calculations to get the final answers for parts a, b, c, and d.