A weight is suspended from a spring and oscillates between positions A and B

[hidemi]

Homework Statement

A weight of mass m is at rest at O when suspended from a spring, as shown. When it is pulled down and released, it oscillates between positions A and B. Which statement about the system consisting of the spring and the mass is correct?

a. The gravitational potential energy of the system is greatest at A.
b. The elastic potential energy of the system is greatest at O.
c. The rate of change of momentum has its greatest magnitude at A and B.
d. The rate of change of gravitational potential energy is smallest at O.
e. The rate of change of gravitational potential energy has its greatest magnitude at A and B.

Ans: C

Relevant Equations

Eq.1 : U = -1/2 * kx^2
Eq.2 : dU/dt = -kx

I just would like to check if my corrections to each of the wrong options are right.
A) & B) elastic potential energy greatest at B, not at A or O, based upon Eq.1
D) Since dU/dt = 0 at O because O is at equilibrium where no change of spring length.
E) At A, dU/dt = -kx. At B, dU/dt = -kx = -kx
Therefore, I think the rate of change of gravitational potential energy is smallest at A and B greatest at O.

I'm not sure if these reasonings are ok?

 

[haruspex]

A bit confusing to use uppercase A, B for points in the diagram and to reference options.

A) & B) elastic potential energy

a asks about GPE.

Since dU/dt = 0 at O because O is at equilibrium where no change of spring length.

Not following your logic. What has this to do with the max rate of change of GPE?

At A, dU/dt = -kx

No, dU/dx = -kx. Everywhere.

 

[hidemi]

A bit confusing to use uppercase A, B for points in the diagram and to reference options.

a asks about GPE.

Not following your logic. What has this to do with the max rate of change of GPE?

No, dU/dx = -kx. Everywhere.

I got it. Sorry for the delayed reply!

 

[Dr.D]

Did you really intend the title to say "A spring is suspended from a spring ..." or perhaps "A mass is suspended ..."?

[Mentor Note -- Thread title fixed now, thanks]

 

[vcsharp2003]

Homework Statement:: A weight of mass m is at rest at O when suspended from a spring, as shown. When it is pulled down and released, it oscillates between positions A and B. Which statement about the system consisting of the spring and the mass is correct?

a. The gravitational potential energy of the system is greatest at A.
b. The elastic potential energy of the system is greatest at O.
c. The rate of change of momentum has its greatest magnitude at A and B.
d. The rate of change of gravitational potential energy is smallest at O.
e. The rate of change of gravitational potential energy has its greatest magnitude at A and B.

Ans: C
Relevant Equations:: Eq.1 : U = -1/2 * kx^2
Eq.2 : dU/dt = -kx

A) & B) elastic potential energy greatest at B, not at A or O, based upon Eq.1

At O, elastic pe = 0 but at all other positions elastic pe = -ve. So shouldn't O be the point where it's max? 0 > (-ve number) always.

 

[haruspex]

At O, elastic pe = 0

No, O is the equilibrium point. The spring is under tension there. The EPE is zero when the spring is relaxed.

all other positions elastic pe = -ve

EPE is never negative, except in a relative sense.

 

[vcsharp2003]

EPE is never negative, except in a relative sense.

If sign is going to be disregarded then option c is the only correct option. Since mass m moves in SHM (simple harmonic motion), so velocity is maximum at equilibrium/mean position and zero at extreme points, whereas acceleration is maximum at extreme points and zero at mean position. O is equilibrium/mean position while A,B are extreme points.

Rate of change of gravitation PE $= \frac {d} {dt} (mgh) = mg \frac {dh} {dt}$, where h is height relative to bottom most point A. At the mean position of O, the velocity magnitude would be maximum i.e. $\frac {dh} {dt}$ would be maximum, so rate of change of gravitation PE would be maximum at equilibrium position O. At A or B extreme points, the rate of change of gravitation PE would be 0 since $\frac {dh} {dt} = 0$ at these points. This means options d and e are ruled out.

Rate of change of momentum $= \frac {d} {dt} (mv) = m \frac {dv} {dt}$, so this quantity is dependent on acceleration of mass m. In SHM, acceleration magnitude is max at extreme points so rate of change of momentum is max at A or B. So, option c is correct.

Gravitational PE is = mgh, where h is height from bottom most point A. So this quantity is greatest at topmost point of B. Option a is therefore ruled out.

Elastic PE is greatest at extreme point of A but not at B, where compression/extension of spring is maximum. At O, elastic PE is greater than 0 since spring is stretched at this position. So, option b is ruled out. At A elastic PE $= \frac {1}{2} k (a +{x_{i}})^2$ and at B elastic PE $= \frac {1}{2} k (a -{x_{i}})^2$, where $x_i$ is the extension of spring when mass m is at equilibrium/mean position and $a$ is the amplitude of vibration about position O i.e. a = OA or OB. Similarly at O elastic PE $= \frac {1}{2} k {x_{i}}^2$.

 

[haruspex]

If sign is going to be disregarded

It's not a question of disregarding signs. EPE cannot be negative, just like KE cannot be negative.

 

[vcsharp2003]

It's not a question of disregarding signs. EPE cannot be negative, just like KE cannot be negative.

The formula for $\text{spring PE} = - \frac {1}{2} k {x}^2$ is not correct since -ve sign is misleading. It should be $\text{spring PE} = \frac {1}{2} k {x}^2$ without the -ve sign.

PE at point 2 in a conservative force field is the work done by an external force in moving the object from position 1 to position 2, where the external force is equal and opposite to the conservative force. Since external force and displacement are in same direction when spring is compressed or extended, so the work done by external force is always positive and therefore elastic PE is always +ve.

Is above explanation correct?

 

[haruspex]

The formula for $\text{spring PE} = - \frac {1}{2} k {x}^2$ is not correct since -ve sign is misleading.

If you are referring to Eq 1 in post #1, you need to know how U is defined there. Looks like it is the work done by the spring in being stretched by x. This is consistent with the force and displacement being measured as positive in the same direction.
This is different from EPE, which refers to the energy stored in the spring, so is -U.

 

[vcsharp2003]

This is consistent with the force and displacement being measured as positive in the same direction.

The spring force and displacement are always in opposite directions. It seems you said that they are in same direction or you meant something else?

 

[haruspex]

The spring force and displacement are always in opposite directions. It seems you said that they are in same direction or you meant something else?

No, I wrote that they are usually taken as positive in the same direction, not that they are in the same direction. Since they will be in opposite directions, one will be positive when the other is negative.
If you take extension as positive then extending the spring will be a positive displacement but the force it exerts will be in the negative direction.