# Conceptual Questions on a Mass Hanging from a Spring

I don't understand what I am doing wrong.

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. Note: the oscillation is small enough that the spring stays stretched beyond its rest length the entire time.

1) Where in the motion is the magnitude of the force from the spring on the mass zero?
Equilibrium, because the change in position delta x is zero.
2) Where in the motion is the magnitude of the net force on the mass a maximum?
The top, because the force of the spring (from compression) and the force of gravity both act on the mass.
3) Where in the motion is the magnitude of the net force on the mass zero?
The bottom, because the force of gravity and the force from the spring oppose each other to keep the block at rest (away from the equilibrium position)
4) Where in the motion is the magnitude of the acceleration a maximum?
The top, because force is maximum there.
5) Where in the motion is the speed zero?
It is zero at the top and bottom; it changes direction at the top and stays at rest at the bottom.
6) Where in the motion is the acceleration zero?
The bottom, since the net force is zero.
7) Where in the motion is the speed a maximum?
At equilibrium.
8) Where in the motion is the magnitude of the force from the spring on the mass a maximum?
The top and bottom, because of delta x.

Yes/No

1) When the object is at half its amplitude from equilibrium, is its speed half its maximum speed?
x(t) = Acos(ωt + φ), cos(ωt + φ) = cos(30 deg) = .5
v(t) = -ωAsin(ωt + φ) = (sqrt(3)/2)Aω, maximum v(t) is Aω
No.
2) When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?
Yes.
a(t) = -(ω^2)Acos(ωt + φ) = .5Aω, maximum a(t) is A(ω^2)

PeterO
Homework Helper
I don't understand what I am doing wrong.

Answer the following questions for a mass that is hanging on a spring and oscillating up and down with simple harmonic motion. Note: the oscillation is small enough that the spring stays stretched beyond its rest length the entire time.

1) Where in the motion is the magnitude of the force from the spring on the mass zero?
Equilibrium, because the change in position delta x is zero.

The spring will apply no force when it is at rest length. It is never at that [see above] so the answer is nowhere

2) Where in the motion is the magnitude of the net force on the mass a maximum?
The top, because the force of the spring (from compression) and the force of gravity both act on the mass.

Net force on an object in SHM is at the extremes, so Top AND Bottom

3) Where in the motion is the magnitude of the net force on the mass zero?
The bottom, because the force of gravity and the force from the spring oppose each other to keep the block at rest (away from the equilibrium position)

Net force is zero when acceleration is minimum, which is at the equilibrium position. Note the spring is always applying an upward force. At the top, gravitational force is bigger, at the bottom spring force is bigger, at middle position they are equal: - net force zero; acceleration zero

4) Where in the motion is the magnitude of the acceleration a maximum?
The top, because force is maximum there.
Also at the bottom

5) Where in the motion is the speed zero?
It is zero at the top and bottom; it changes direction at the top and stays at rest at the bottom.
It actually only changes direction at the bottom too

6) Where in the motion is the acceleration zero?
The bottom, since the net force is zero.

AT equilibrium because there net force is zero

7) Where in the motion is the speed a maximum?
At equilibrium.

Yay - true

8) Where in the motion is the magnitude of the force from the spring on the mass a maximum?
The top and bottom, because of delta x.

Spring pulls most strongly when it is stretched the most - so at the bottom only

Yes/No

1) When the object is at half its amplitude from equilibrium, is its speed half its maximum speed?
x(t) = Acos(ωt + φ), cos(ωt + φ) = cos(30 deg) = .5
v(t) = -ωAsin(ωt + φ) = (sqrt(3)/2)Aω, maximum v(t) is Aω
No.

true

2) When the object is at half its amplitude from equilibrium, is the magnitude of its acceleration at half its maximum value?
Yes.
a(t) = -(ω^2)Acos(ωt + φ) = .5Aω, maximum a(t) is A(ω^2)

true

Thanks, I see what I did wrong now.