1) A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harmonic motion with an amplitude of 10 cm. What is the speed of the 1.0 kg mass at the equilibrium position?
w = .1 x sqrt(400/1) = 2 m/s
To find the speed of the 1.0 kg mass at the equilibrium position in simple harmonic motion, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system, consisting of the mass and the spring, remains constant throughout the motion.
The total mechanical energy is given by the sum of the potential energy (PE) and the kinetic energy (KE). At the equilibrium position, the potential energy is at its minimum, and the kinetic energy is at its maximum.
To find the speed at the equilibrium position, we first need to calculate the potential energy at the equilibrium position. The potential energy of a spring is given by the formula:
PE = (1/2)kx^2
Where k is the force constant of the spring and x is the displacement from the equilibrium position. At the equilibrium position, the displacement is zero, so the potential energy is also zero.
Next, we can find the kinetic energy at the equilibrium position. The kinetic energy is given by the formula:
KE = (1/2)mv^2
Where m is the mass of the object and v is its velocity. At the equilibrium position, the object has its maximum velocity, and the displacement is zero. Therefore, the kinetic energy is at its maximum.
Since the potential energy is zero at the equilibrium position, the total mechanical energy is equal to the kinetic energy at the equilibrium position. Substituting the values into the equation:
KE = (1/2)mv^2
We get:
(1/2)mv^2 = (1/2)kx^2
Substituting the given values:
(1/2)(1 kg)v^2 = (1/2)(400 N/m)(0.1 m)^2
Simplifying:
v^2 = (400 N/m)(0.1 m)^2
v^2 = 4 N m
Taking the square root of both sides:
v = 2 m/s
Therefore, the speed of the 1.0 kg mass at the equilibrium position is 2 m/s.