A spring with a weight attached is oscillating. The weight (which is attached at the bottom of the spring) is 5 feet from a 10-foot ceiling when it’s at rest. The motion of the weight can be described by the equation: y=3sin(pi*t - pi/2) , where y is the distance from the equilibrium point after time t (in seconds).
a) Was the weight pulled down or pushed up before it was released?
b) How far was the weight from the ceiling when it was released?
c) How close will the weight come to the ceiling?
d) When does the weight first pass it’s equilibrium point?
e) What is the greatest distance that the weight will be from the ceiling?
f) Find the period of the motion. g) Find the amplitude of the motion.
h) What is the frequency of the motion?
i) How far from the ceiling is the weight after 2.5 seconds?
If you have no ideas on any of these questions, you seriously need to review your trig functions.
y = 5+3sin(πt-π/2) = 5+3sin(π(t-1/2))
(a) at t=0, y = 3sin(-π/2) = -3
Since that is below the equilibrium point, the weight was pulled down.
(b) all these distance from the ceiling questions can be answered by considering that the distance is just d = 5-y
(f) since the period of sin(kt) is 2π/k, we have a period of 2π/π = 2 seconds.
(g) the amplitude is clearly 3.
(h) f = 1/p
So, what do you get for the others, and how do you get it?
a) To determine whether the weight was pulled down or pushed up before it was released, we need to examine the equation for the motion:
y = 3sin(pi*t - pi/2)
In this equation, the coefficient of sine, which is 3, represents the amplitude of the motion. Since the weight is at rest at y = 5 feet (5 units), we can conclude that it was pulled down before being released. The negative displacement indicates a downward direction.
b) To find how far the weight was from the ceiling when it was released, we need to evaluate the equation at t = 0:
y = 3sin(pi*0 - pi/2)
y = 3sin(-pi/2)
y = 3(-1)
y = -3
Therefore, the weight was 3 feet from the ceiling when it was released.
c) To determine how close the weight will come to the ceiling, we need to analyze the amplitude of the motion. From the equation, we know that the amplitude is 3. Thus, the weight will get as close to the ceiling as 3 + 5 = 8 feet.
d) The weight would first pass its equilibrium point when y = 0. We can solve for t by setting y = 0 in the equation and solving for t:
0 = 3sin(pi*t - pi/2)
sin(pi*t - pi/2) = 0
To find the first value of t that satisfies this equation, we look for the first value where sin equals 0. This occurs when pi*t - pi/2 = 0 or when pi*t = pi/2. Therefore:
pi*t = pi/2
t = 1/2 seconds
Therefore, the weight first passes its equilibrium point at t = 1/2 seconds.
e) The greatest distance that the weight will be from the ceiling is equal to the amplitude of the motion. From the equation, we know that the amplitude is 3 feet.
f) The period of the motion can be determined from the equation y = 3sin(pi*t - pi/2). The coefficient in front of t in the argument of sin is pi. The period (T) is given by T = 2*pi/abs(pi) = 2. Therefore, the period of the motion is 2 seconds.
g) The amplitude of the motion is given by the coefficient in front of sin in the equation. In this case, the amplitude is 3 feet.
h) The frequency of the motion is the reciprocal of the period, which is 1/T. From part f), we determined that the period is 2 seconds. Therefore, the frequency is 1/2 Hz.
i) To find how far from the ceiling the weight is after 2.5 seconds, we substitute t = 2.5 into the equation:
y = 3sin(pi*2.5 - pi/2)
y = 3sin(2*pi - pi/2)
y = 3sin(3*pi/2)
y = 3(-1)
y = -3
Therefore, the weight is 3 feet from the ceiling after 2.5 seconds.
a) Since the weight is 5 feet from the ceiling when it's at rest, we can conclude that the weight was pulled down before it was released.
b) When the weight is at rest, it is 5 feet from the ceiling. Therefore, the distance from the ceiling when it was released is also 5 feet.
c) The weight oscillates according to the equation y = 3sin(pi*t - pi/2). The maximum displacement from the equilibrium point is given by the amplitude of the equation, which is 3. Therefore, the weight will come closest to the ceiling at a distance of 5 - 3 = 2 feet.
d) The weight passes its equilibrium point when y = 0. Solving the equation 3sin(pi*t - pi/2) = 0, we get t = 0 and t = 1. Therefore, the weight first passes its equilibrium point at t = 0 and t = 1.
e) The greatest distance that the weight will be from the ceiling is given by the amplitude of the equation, which is 3 feet.
f) The period of the motion is the time it takes for one complete cycle. In this case, the equation is y = 3sin(pi*t - pi/2), which means the motion completes a cycle when the argument of the sine function, (pi*t - pi/2), increases by 2pi. So the period is 2pi/pi = 2 seconds.
g) The amplitude of the motion is the maximum displacement from the equilibrium point, which is 3 feet.
h) The frequency of the motion is the number of cycles per unit of time. In this case, the period is 2 seconds, so the frequency is 1/2 = 0.5 cycles per second.
i) To find how far from the ceiling the weight is after 2.5 seconds, we can substitute t = 2.5 into the equation y = 3sin(pi*t - pi/2):
y = 3sin(pi*2.5 - pi/2)
y = 3sin(7.5pi - pi/2)
Calculating the sine of this angle will give us the distance from the equilibrium point after 2.5 seconds.