A particle executes simple harmonic motion such that at t = 0 it is at the amplitude of oscillation A = 22.5 cm. The period of the oscillation is 0.25 s.
When is the first time this particle will be at x = -1/2 the amplitude, moving away from equilibrium?
To find the first time the particle will be at x = -1/2 the amplitude, moving away from equilibrium, we need to determine the phase of the particle's motion at that position.
Here's how we can solve the problem:
1. Determine the angular frequency (ω) of the simple harmonic motion using the formula:
ω = 2π / T
where T is the period of oscillation. In this case, T = 0.25 s, so:
ω = 2π / 0.25 = 8π rad/s
2. The equation of motion for simple harmonic motion can be written as:
x = A * cos(ωt + φ)
where x is the displacement, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.
3. At t = 0, the particle is at the amplitude of oscillation A = 22.5 cm. Substituting these values into the equation of motion, we get:
22.5 cm = A * cos(0 + φ)
22.5 cm = A * cos(φ)
Since cos(φ) = 1 when φ = 0, we can rewrite the equation as:
22.5 cm = A * 1
A = 22.5 cm
4. Now we want to find the phase constant φ when x = -1/2 the amplitude, which is -11.25 cm. Substituting these values into the equation of motion, we get:
-11.25 cm = A * cos(ωt + φ)
-11.25 cm = 22.5 cm * cos(ωt + φ)
Dividing both sides by 22.5 cm, we have:
-0.5 = cos(ωt + φ)
5. To find the phase constant φ, we need to take the inverse cosine (also known as arccos) of -0.5. Using a calculator, we find:
φ ≈ 2.0944 rad
6. Now we can find the time when the particle reaches x = -1/2 the amplitude, moving away from equilibrium. Rearranging the equation of motion, we have:
cos(ωt + φ) = -0.5
Taking the inverse cosine of both sides, we get:
ωt + φ = arccos(-0.5)
Subtracting φ from both sides, we have:
ωt = arccos(-0.5) - φ
Finally, dividing both sides by ω, we obtain:
t = (arccos(-0.5) - φ) / ω
Plugging in the values, we have:
t = (arccos(-0.5) - 2.0944) / 8π
Evaluating this expression using a calculator, we find:
t ≈ 0.044 s
Therefore, the first time the particle will be at x = -1/2 the amplitude, moving away from equilibrium, is approximately 0.044 seconds.