A 2.00 kg frictionless block is attached to a horizontal spring as shown. At t = 0, the position x = 0.225 m, and the velocity is 4.25 m/s toward the right in the positive x direction. Position x as a function of t is: x = A*cos(ωt ) , where A is the amplitude of motion and ω is the angular frequency discussed Chapter 11 and the notes.
(a) (3 points) What is the spring constant k?
(b) (25 points) Use conservation of energy to compute amplitude A.
(c) (5 points) How much farther from the point shown will the block move before it momentarily comes to rest before turning around?
(d) (4 points) What is the period T of the motion?
(e) (3 points) If the mass of this problem was doubled to 4.00 kg, how would your answer to part (d ) change?
(a) To find the spring constant k, we need to use Hooke's law. Hooke's law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
From the given information, we know the mass m = 2.00 kg, and at t = 0, the position x = 0.225 m. We also know that the block is frictionless, so there is no other force acting on it.
At the equilibrium position, the force from the spring must balance the weight of the block. The weight is given by W = m * g, where g is the acceleration due to gravity.
Since there is no vertical motion, the force from the spring is k * x, where x is the displacement from the equilibrium position.
Therefore, we have:
k * x = m * g
Plugging in the values, we get:
k * 0.225 m = 2.00 kg * 9.8 m/s^2
Solving for k, we have:
k = (2.00 kg * 9.8 m/s^2) / 0.225 m
Calculating the result, we find:
k ≈ 87.11 N/m
Therefore, the spring constant k is approximately 87.11 N/m.
(b) To compute the amplitude A using conservation of energy, we can use the fact that the total mechanical energy of the system is conserved.
The total mechanical energy is given by the sum of the kinetic energy and the potential energy. At t = 0, the block has a velocity of 4.25 m/s.
The kinetic energy is given by KE = (1/2) * m * v^2.
The potential energy stored in the spring is given by PE = (1/2) * k * x^2.
Since the block is initially at its maximum displacement, the potential energy is equal to the total mechanical energy.
Therefore:
KE = PE
(1/2) * m * v^2 = (1/2) * k * x^2
We can substitute the values into the equation:
(1/2) * 2.00 kg * (4.25 m/s)^2 = (1/2) * (87.11 N/m) * (0.225 m)^2
Simplifying the equation, we find:
(1/2) * 2.00 kg * 18.06 m^2/s^2 = (1/2) * (87.11 N/m) * 0.050625 m^2
36.12 J = 2.559 J * A^2
Solving for A, we have:
A^2 = 36.12 J / 2.559 J
A^2 ≈ 14.11
Taking the square root, we have:
A ≈ √14.11 ≈ 3.75 m
Therefore, the amplitude A is approximately 3.75 m.
(c) To find how much farther from the point shown the block moves before it momentarily comes to rest before turning around, we need to find the maximum displacement from the equilibrium position.
The maximum displacement occurs when the block momentarily comes to rest before turning around. At this point, the velocity is momentarily zero.
Using the equation for displacement as a function of time:
x = A * cos(ωt)
We can find the time t when the velocity is zero:
v = -A * ω * sin(ωt) = 0
sin(ωt) = 0
This equation is true when ωt = n * π, where n is an integer. Since the block is already at a displacement of 0.225 m, we can find the maximum displacement:
x_max = A + 0.225 m
At ωt = π, the block is at its maximum displacement in the positive x direction. Therefore, we have:
ω * π = π
Solving for ω, we find:
ω = 1 rad/s
Substituting the values into the equation, we have:
x_max = 3.75 m + 0.225 m
x_max = 3.975 m
Therefore, the block moves 3.975 m farther from the point shown before it momentarily comes to rest before turning around.
(d) The period T of the motion is the time taken for one complete cycle. In other words, it is the time taken for the block to return to its initial position and velocity.
The period of motion is given by T = 2π/ω, where ω is the angular frequency.
From part (a), we found that ω = 1 rad/s. Substituting the value into the equation, we have:
T = 2π/1 rad/s
Simplifying the equation, we find:
T = 2π s ≈ 6.28 s
Therefore, the period T of the motion is approximately 6.28 s.
(e) If the mass of this problem was doubled to 4.00 kg, the period T of the motion would not change.
The period of motion depends only on the angular frequency ω, which is determined by the spring constant k and the mass m. The mass does not appear in the equation for the period.
Therefore, doubling the mass to 4.00 kg would not change the period T of the motion.
(a) To find the spring constant k, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, F = -kx, where F is the force, x is the displacement, and k is the spring constant.
In this problem, we can determine the spring constant by analyzing the given information. At t = 0, the position x = 0.225 m, and the velocity is 4.25 m/s toward the right. Since the block is at its maximum displacement and moving in the positive x direction, the force acting on it must be directed toward the equilibrium position (opposite to the displacement). Hence, the force is given by F = -kx.
Using the given values, we can plug them into the equation:
F = -kx
-(-k)(0.225) = 2.00 * 4.25
Simplifying the equation, we get:
k = (2.00 * 4.25) / 0.225
k = 38.00 N/m
Therefore, the spring constant k is 38.00 N/m.
(b) To compute the amplitude A using conservation of energy, we need to recognize that the spring-block system undergoes simple harmonic motion. Simple harmonic motion involves the exchange between kinetic energy (KE) and potential energy (PE) in a system.
At t = 0, the block's position x is given as 0.225 m, and its velocity is given as 4.25 m/s toward the right. We can find the total mechanical energy (E) at this point, which remains constant throughout the motion:
E = KE + PE
The kinetic energy is given by:
KE = (1/2) * m * v^2
Substituting the given values, we find:
KE = (1/2) * 2.00 * (4.25)^2
KE = 19.03125 J
The potential energy of a spring is given by:
PE = (1/2) * k * x^2
Substituting the given values, we have:
PE = (1/2) * 38.00 * (0.225)^2
PE = 0.95625 J
Since the mechanical energy remains constant, we can equate the total mechanical energy at t = 0 to the total mechanical energy at any other point in time:
E = KE + PE
E = 19.03125 + 0.95625
E = 19.9875 J
The amplitude A can be calculated by equating the total mechanical energy to the potential energy equation:
19.9875 = (1/2) * k * A^2
Solving for A, we get:
A^2 = (2 * 19.9875) / 38.00
A^2 = 1.05290131579
Taking the square root of both sides, we find:
A = 1.026 m
Therefore, the amplitude A is 1.026 m.
(c) To determine how much farther the block moves before it momentarily comes to rest before turning around, we need to find the maximum displacement from the equilibrium position. This occurs when the block momentarily comes to rest before changing direction.
The maximum displacement can be found using the amplitude A, which is given as 1.026 m. Since the block is currently at x = 0.225 m, the additional displacement needed can be calculated as:
Additional displacement = A - x
Additional displacement = 1.026 - 0.225
Additional displacement = 0.801 m
Therefore, the block will move an additional 0.801 m in the positive x direction before momentarily coming to rest.
(d) The period T of the motion can be determined from the angular frequency ω. The equation x = A * cos(ωt) relates the position x of the block to the angular frequency ω and time t.
Comparing this equation to the one given, we can see that ω = k/m. We have previously calculated that the spring constant k is 38.00 N/m, and the mass m is 2.00 kg.
Substituting the values into the equation, we get:
ω = 38.00 / 2.00
ω = 19.00 rad/s
The period T of motion is the time taken for one complete cycle, which is given by:
T = 2π / ω
T = 2π / 19.00
T = 0.33 s
Therefore, the period T of the motion is 0.33 s.
(e) If the mass of the system is doubled to 4.00 kg, the period T of the motion will change. The period of an object undergoing simple harmonic motion is independent of its mass, meaning it remains the same regardless of the mass.
Therefore, the answer to part (d) remains unchanged, and the period T of the motion remains 0.33 s.