The position of a 0.43 kg object attached to a spring is described by [x = (0.49 m) cos(1.25πt)]
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 = 0.01 s.
d) What is in meters/seconds2 the maximum acceleration of the oscillating object?
To answer these questions, we need to use the given equation of motion for the object attached to the spring, which is given by:
x = (0.49 m) * cos(1.25πt)
a) The amplitude of oscillations represents the maximum distance the object moves from its equilibrium position. In this case, the amplitude is equal to the coefficient of the cosine function, which is 0.49 m. Therefore, the amplitude of oscillations is 0.49 m.
b) The maximum speed of the oscillating object occurs when the object passes through its equilibrium position. At this point, the velocity reaches a maximum value. The maximum velocity is given by the derivative of x with respect to t:
v_max = dx/dt = - (0.49 m) * sin(1.25πt) * (1.25π)
To find the maximum speed, substitute t = 0 into the equation:
v_max = - (0.49 m) * sin(1.25π * 0) * (1.25π)
= 0
Therefore, the maximum speed of the oscillating object is 0 m/s.
c) To find the object's speed at t = 0.01 s, we need to substitute t = 0.01 into the equation for velocity:
v = - (0.49 m) * sin(1.25π * 0.01) * (1.25π)
Calculating this expression will give you the speed in meters/second at t = 0.01 s.
d) The maximum acceleration of the oscillating object occurs when the object is at the extremes of its motion, that is, when x = ±0.49 m. At these points, the acceleration reaches its maximum value. The maximum acceleration is given by the derivative of velocity with respect to time:
a_max = dv/dt = - (0.49 m) * cos(1.25πt) * (1.25π)
To find the maximum acceleration, substitute t = 0 into the equation:
a_max = - (0.49 m) * cos(1.25π * 0) * (1.25π)
= - (0.49 m) * (1.25π)
Therefore, the maximum acceleration of the oscillating object is approximately -1.542 m/s².
To find the answers to the given questions, we need to understand the equation that describes the object's position and use it to derive the necessary information. The equation provided is:
x = (0.49 m) cos(1.25πt)
a) To find the amplitude of oscillations, we need to use the cosine function. The general form of a cosine function is given by:
y = A cos(Bx)
Where A represents the amplitude. By comparing this general form to our equation, we can see that the amplitude is 0.49 m. Therefore, the amplitude of oscillations is 0.49 meters.
b) The maximum speed of the oscillating object is equal to the magnitude of the maximum value of the derivative of the position function with respect to time. In other words, we need to find the maximum value of dx/dt.
To find this, we differentiate the position function with respect to time (t):
x = (0.49 m) cos(1.25πt)
dx/dt = -(0.49 m) (1.25π sin(1.25πt))
To find the maximum value, we evaluate the derivative at the critical points. In this case, the critical point occurs when sin(1.25πt) is either 1 or -1. So, we have:
dx/dt = -(0.49 m) (1.25π)
Using the given values, we can calculate the maximum speed:
Speed = |dx/dt| = |-(0.49 m) (1.25π)| = 0.61 m/s
Therefore, the maximum speed of the oscillating object is 0.61 meters/second.
c) To find the object's speed at t = 0.01 s, we substitute the given time value into the derivative equation we derived earlier:
dx/dt = -(0.49 m) (1.25π sin(1.25πt))
At t = 0.01 s:
Speed = |-(0.49 m) (1.25π sin(1.25π(0.01)))|
By evaluating this expression, we can find the object's speed at t = 0.01 s.
d) The maximum acceleration of the oscillating object is equal to the magnitude of the maximum value of the second derivative of the position function with respect to time. In other words, we need to find the maximum value of d²x/dt².
To find this, we differentiate the derivative we found earlier with respect to time (t):
d²x/dt² = -(0.49 m) (1.25π)² cos(1.25πt)
The maximum acceleration occurs when cos(1.25πt) is either 1 or -1. So, we have:
d²x/dt² = -(0.49 m) (1.25π)²
Using the given values, we can calculate the maximum acceleration:
Acceleration = |d²x/dt²| = |-(0.49 m) (1.25π)²| = 9.4 m/s²
Therefore, the maximum acceleration of the oscillating object is 9.4 meters/second².