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A block with mass 1.793 kg sliding on a frictionless surface is attached to a horizontal spring with spring constant 11.79 N/m at the equilibrium position of the spring. The block is then pulled a distance 13.61 cm from the equilibrium point and released from rest. At what time is the block located a distance of 12.21 cm from the equilibrium point?

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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.

User Jan Misker
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Final answer:

The block will be located a distance of 12.21 cm from the equilibrium point at a time of 0.283 seconds.

Step-by-step explanation:

The time it takes for the block to be located a distance of 12.21 cm from the equilibrium point can be found using the concept of simple harmonic motion (SHM). In SHM, the displacement of the block from the equilibrium point can be described by the equation: x(t) = A * cos(ωt), where A is the amplitude of the motion, ω is the angular frequency, and t is time.

In this case, the amplitude A is the initial displacement of the block from the equilibrium point, which is 13.61 cm. The angular frequency ω can be calculated using the formula: ω = sqrt(k/m), where k is the spring constant and m is the mass of the block.

Plugging in the given values, we have:
A = 13.61 cm = 0.1361 m
k = 11.79 N/m
m = 1.793 kg

Using the formula for the angular frequency, we find:
ω = sqrt(11.79 N/m / 1.793 kg) = 2.2824 rad/s

To find the time at which the block is located a distance of 12.21 cm from the equilibrium point, we can rearrange the equation for SHM:
t = (1/ω) * acos(x/A)

Plugging in the given values, we have:
x = 12.21 cm = 0.1221 m
A = 0.1361 m
ω = 2.2824 rad/s

Calculating the time, we find:
t = (1/2.2824) * acos(0.1221/0.1361) = 0.283 s

User Sqwerl
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