To tackle the problem involving the particle's displacement equation, we need to analyze the given equation: y = 3sin((π/3)t + π/6) cm. This equation describes simple harmonic motion (SHM), where the particle oscillates back and forth. Let's break down the steps to find the required times for when the particle comes to rest and when it has maximum velocity and acceleration.
Understanding the Displacement Equation
The displacement equation can be expressed in the form of y = A sin(ωt + φ), where:
- A = amplitude (3 cm in this case)
- ω = angular frequency (π/3 rad/s)
- φ = phase constant (π/6 rad)
Finding Times When the Particle Comes to Rest
A particle in SHM comes to rest when its velocity is zero. The velocity (v) can be derived from the displacement equation by differentiating it with respect to time (t):
v = dy/dt = Aω cos(ωt + φ)
Substituting the values:
v = 3 * (π/3) * cos((π/3)t + π/6) = π cos((π/3)t + π/6)
To find when the particle comes to rest, we set the velocity to zero:
π cos((π/3)t + π/6) = 0
This occurs when cos((π/3)t + π/6) = 0. The cosine function is zero at odd multiples of π/2:
(π/3)t + π/6 = (2n + 1)π/2, where n is an integer.
Solving for Time
Rearranging gives:
(π/3)t = (2n + 1)π/2 - π/6
t = (3/π) * [(2n + 1)π/2 - π/6]
t = (3/π) * [(2n + 1)π/2 - π/6] = (3/π) * [(2n + 1)π/2 - π/6]
t = (3/π) * [(2n + 1)π/2 - π/6]
Now, let's calculate the specific times for n = 0, 1, 2, etc., to find the 2nd, 4th, and 5th times:
- For n = 0: t = (3/π) * [π/2 - π/6] = (3/π) * [3π/6 - π/6] = (3/π) * (2π/6) = 1 s
- For n = 1: t = (3/π) * [3π/2 - π/6] = (3/π) * [9π/6 - π/6] = (3/π) * (8π/6) = 4 s
- For n = 2: t = (3/π) * [5π/2 - π/6] = (3/π) * [15π/6 - π/6] = (3/π) * (14π/6) = 7 s
Determining Times for Maximum Velocity
Maximum velocity occurs when the cosine function equals ±1. Thus, we set:
cos((π/3)t + π/6) = ±1
This happens at even multiples of π:
(π/3)t + π/6 = kπ, where k is an integer.
Solving for Time
Rearranging gives:
(π/3)t = kπ - π/6
t = (3/π)(kπ - π/6) = 3k - 0.5
Now, we find the specific times for k = 1, 2, 3, etc., to get the 3rd, 5th, and 6th times:
- For k = 1: t = 3(1) - 0.5 = 2.5 s
- For k = 2: t = 3(2) - 0.5 = 5.5 s
- For k = 3: t = 3(3) - 0.5 = 8.5 s
Finding Times for Maximum Acceleration
Maximum acceleration occurs when the sine function is at its maximum or minimum, which is ±A. The acceleration (a) can be derived as:
a = d²y/dt² = -Aω² sin(ωt + φ)
Setting sin(ωt + φ) = ±1 gives us the conditions for maximum acceleration:
ωt + φ = nπ + π/2, where n is an integer.
Solving for Time
Rearranging gives:
t = (nπ + π/2 - φ)/ω
Substituting the values:
t = (nπ + π/2 - π/6)/(π/3) = 3(nπ + π/2 - π/6)/π = 3n + 1.5 - 0.5 = 3n + 1
Now, we find the specific times for n = 2, 3, 4, etc., to get the 3rd, 4th, and 6th times:
- For n = 2: t = 3(2) + 1 = 7 s
- For n = 3: t = 3(3) + 1 = 10 s
- For n = 4: t = 3(4) + 1 = 13 s
In summary, we have calculated the times for the particle's motion based on its displacement equation. The particle comes to rest at 1 s, 4 s, and 7 s; it achieves maximum velocity at 2.5 s, 5.5 s
