Question

As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change

Answer 1

As a mass on a spring moves farther from the equilibrium position, the velocity, acceleration, and force change.

Step-by-step explanation:

As a mass on a spring moves farther from the equilibrium position, the velocity, acceleration, and force change.

Velocity: As the mass moves away from the equilibrium position, its velocity increases. The mass reaches its maximum velocity at the maximum displacement.

Acceleration: The acceleration of the mass is greatest at the equilibrium position and decreases as the mass moves away from it. The acceleration is always directed towards the equilibrium position.

Force: The force exerted by the spring is directly proportional to the displacement from the equilibrium position. As the mass moves farther from the equilibrium position, the force exerted by the spring increases.

Answer 2

Refer to the diagram shown below.

m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
v is maximum;
a is zero.

At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.