Step-by-step explanation:
To find the speed of the mass at t = 2.50 s, we can use the equation for simple harmonic motion:
x(t) = A * cos(ωt + φ)
where:
x(t) = displacement of the mass from equilibrium at time t
A = amplitude of the oscillation
ω = angular frequency of the oscillation
φ = phase angle (initial phase)
Given that the period T = 4.60 s, we know that the angular frequency is related to the period as follows:
ω = 2π / T
Substitute the given period T = 4.60 s to find ω:
ω = 2π / 4.60 s ≈ 1.363 s^(-1)
At t = 0 s, the mass has zero speed and is at x = 8.30 cm. This means that the initial phase angle φ = 0.
Now, we can find the displacement x at t = 2.50 s:
x(2.50 s) = A * cos(ω * 2.50 s)
We know that at t = 0 s, x = 8.30 cm, so:
8.30 cm = A * cos(0)
Since the cosine of 0 is 1, we can find the amplitude A:
A = 8.30 cm
Now, calculate the displacement at t = 2.50 s:
x(2.50 s) = 8.30 cm * cos(1.363 s^(-1) * 2.50 s) ≈ 8.30 cm * cos(3.4075) ≈ 3.06 cm
The displacement at t = 2.50 s is approximately 3.06 cm. However, the speed is the magnitude of the derivative of the displacement with respect to time:
v(t) = d(x(t))/dt = -A * ω * sin(ωt)
Substitute the known values:
v(2.50 s) = -8.30 cm * 1.363 s^(-1) * sin(1.363 s^(-1) * 2.50 s) ≈ -8.30 cm * 1.363 s^(-1) * sin(3.4075) ≈ -8.30 cm * 1.363 s^(-1) * (-0.2357) ≈ 3.06 cm/s
So, the speed of the mass at t = 2.50 s is approximately 3.06 cm/s.