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?

Maximum kinetic energy (which occurs at the equilbrium position) equals maximum stored spring potential energy.

Therefore

(1/2) M Vmax^2 = (1/2) k X^2

Vmax = X*sqrt(k/M)

X is the amplitude, 0.100 m.

To find the speed of the mass at the equilibrium position, we can use the principles of simple harmonic motion.

First, let's understand a few key concepts:
1. In simple harmonic motion, the object oscillates back and forth around an equilibrium position.
2. The equilibrium position is the point where the net force on the mass is zero. In this case, it corresponds to the relaxed position of the spring.
3. The amplitude of the motion is the maximum displacement from the equilibrium position. In this case, it's given as 10 cm.

Given that the amplitude is 10 cm, we can convert it to meters by dividing by 100: 10 cm / 100 = 0.10 m.

The maximum potential energy of the mass occurs when it is at the extreme points of the motion, which is equal to the maximum elastic potential energy stored in the spring. The potential energy can be calculated using the formula:

Potential energy (PE) = (1/2) * k * x^2

where k is the force constant of the spring and x is the displacement from the equilibrium position.

In this case, x is equal to the amplitude, so x = 0.10 m.

Substituting the given values, we can calculate the potential energy:

PE = (1/2) * 400 N/m * (0.10 m)^2
= (1/2) * 400 N/m * 0.01 m^2
= 2 J

The potential energy at the equilibrium position is zero, so all of this potential energy is converted to kinetic energy at this point. Therefore, the kinetic energy at the equilibrium position is equal to 2 J.

The kinetic energy (KE) can be calculated using the formula:

Kinetic energy (KE) = (1/2) * m * v^2

where m is the mass and v is the velocity.

In this case, m = 1.0 kg.

Substituting the given values, we can solve for v:

2 J = (1/2) * 1.0 kg * v^2

Dividing both sides by (1/2) * 1.0 kg:

4 J / 1.0 kg = v^2

Simplifying:

4 m^2/s^2 = v^2

Taking the square root of both sides:

2 m/s = v

Therefore, the speed of the 1.0 kg mass at the equilibrium position is 2 m/s.