Final answer:
To find the time at which the block is located a distance of 12.21 cm from the equilibrium point, we can use the equation of motion for simple harmonic motion. First, we find the angular frequency using the spring constant and mass of the block. Then, we find the phase angle using the initial displacement and amplitude of the motion. Finally, we solve for the time when the block is located at the given distance from the equilibrium point.
Step-by-step explanation:
To find the time at which the block is located a distance of 12.21 cm from the equilibrium point, we can use the equation of motion for simple harmonic motion (SHM):
x(t) = A * cos(ωt + φ)
where x(t) is the position of the block at time t, A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle.
First, let's find the angular frequency ω:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the block.
Substituting the given values:
ω = sqrt(11.79 N/m / 1.793 kg)
ω ≈ 2.443 rad/s
Next, let's find the phase angle φ:
φ = cos^(-1)(x₀/A)
where x₀ is the initial displacement of the block from the equilibrium point and A is the amplitude of the motion.
Substituting the given values:
φ = cos^(-1)(0.1361 m / 0.1361 m)
φ ≈ 0 rad
Finally, we can find the time t when the block is located at a distance of 12.21 cm from the equilibrium point:
x(t) = 0.1221 m = 0.1361 m * cos(2.443 rad/s * t)
To solve for t, we can rearrange the equation:
cos^(-1)(0.1221 m / 0.1361 m) = 2.443 rad/s * t
t ≈ cos^(-1)(0.897) / (2.443 rad/s)
t ≈ 0.544 s
Therefore, the block is located a distance of 12.21 cm from the equilibrium point at approximately 0.544 seconds.