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?

To find the speed of the 1.0 kg mass at the equilibrium position, we need to apply the concepts of simple harmonic motion.

The equilibrium position is the point at which the spring is neither stretched nor compressed. In this case, it is where the mass comes to rest when it is not in motion.

When an object is at its equilibrium position in simple harmonic motion, its speed is at its maximum. This is because the object is transitioning from motion in one direction to motion in the opposite direction, temporarily halting at zero velocity.

To find the speed at the equilibrium position, we can use the equation for simple harmonic motion:

v = Aω

where
v is the speed,
A is the amplitude of the motion, and
ω (omega) is the angular frequency of the motion.

The angular frequency, ω, can be calculated using

ω = √(k/m)

where
k is the force constant (spring constant),
m is the mass attached to the spring.

Here, k = 400 N/m and m = 1.0 kg.

Substituting these values into the equation, we get:

ω = √(400 N/m / 1.0 kg)

ω = √(400 rad^2/s^2)

ω = 20 rad/s

Now, we can substitute the value of the amplitude (A) and the angular frequency (ω) into the equation for speed:

v = Aω

v = (0.1 m) * (20 rad/s)

v = 2 m/s

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