Answer:
0.835cm or 1.145cm
Step-by-step explanation:
We know that in simple harmonic motion, the speed is at its maximum at the equilibrium point (midpoint) and zero at the amplitude. Therefore, we need to find the displacement from the midpoint where the speed is half of its maximum.
Let's start by finding the maximum velocity. We know that the velocity is given by:
v = Aωcos(ωt)
where A is the amplitude, ω is the angular frequency, and t is the time. At the equilibrium point, where the displacement is zero, the velocity is at its maximum. Therefore:
v_max = Aω
Next, we need to find the velocity when the speed is half of v_max. The speed is given by the absolute value of the velocity:
speed = |v| = Aω|cos(ωt)|
When the speed is half of v_max, we have:
Aω|cos(ωt)| = 0.5v_max
Substituting v_max = Aω, we get:
|cos(ωt)| = 0.5
Since the cosine function oscillates between -1 and 1, we have two possible solutions:
cos(ωt) = 0.5 or cos(ωt) = -0.5
Solving for ωt, we get:
ωt = arccos(0.5) = π/3 or ωt = 2π/3
or
ωt = -arccos(0.5) = -π/3 or ωt = -2π/3
We only need to consider the positive solutions, since displacement is always positive. Therefore:
ωt = π/3 or ωt = 2π/3
The displacement corresponding to these times can be found using the equation for displacement in simple harmonic motion:
x = Acos(ωt)
Substituting ωt = π/3, we get:
x = 1.67cos(π/3) = 0.835 cm
Substituting ωt = 2π/3, we get:
x = 1.67cos(2π/3) = 1.145 cm
Therefore, the particle's speed equals one half of its maximum speed at a positive displacement of either 0.835 cm or 1.145 cm from the midpoint of its motion.