What is the amplitude of the motion? Remember, at t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. Enter your answer as a numerical value.
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To find the amplitude of the motion, we need to understand the characteristics of the system and the motion of the block attached to the spring.
In a simple harmonic motion, the amplitude represents the maximum displacement from the equilibrium position. It is the distance between the extreme positions of the block during its oscillation.
Given that the block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -0.33 m from the equilibrium position of the spring, we can derive the amplitude by using the energy conservation principle.
The energy of a block-spring system can be expressed as the sum of kinetic energy and potential energy:
E = KE + PE
The kinetic energy (KE) can be calculated using the formula:
KE = 1/2 * M * v^2
where M is the mass of the block and v is the velocity.
The potential energy (PE) of a spring can be calculated using the formula:
PE = 1/2 * k * x^2
where k is the spring constant and x is the displacement from the equilibrium position.
Since the energy is conserved, the total energy E should remain constant throughout the motion.
E_initial = E_final
Therefore,
KE_initial + PE_initial = KE_final + PE_final
Since the block starts from a position of xo = -0.33 m, where the potential energy is at its maximum, the initial potential energy is:
PE_initial = 1/2 * k * xo^2
Similarly, the final potential energy is zero when the block reaches its extreme positions.
PE_final = 0
So, the energy equation simplifies to:
KE_initial = KE_final
Now let's substitute the formulas for kinetic energy and solve for the velocity (v) at the equilibrium position:
1/2 * M * v^2 = 1/2 * k * xo^2
Simplifying,
v^2 = k * xo^2 / M
v = sqrt(k * xo^2 / M)
Substituting the given values:
v = sqrt(61.2 N/m * (-0.33 m)^2 / 5 kg)
v ≈ -0.659 m/s (taking the negative sign because the block is originally moving in the negative direction)
Now we have the velocity (v) at the equilibrium position, and the amplitude (A) is the absolute value of the displacement at this position:
A = |xo| = |-0.33 m| = 0.33 m
Therefore, the amplitude of the motion is approximately 0.33 m.