Springs - AP Physics 1

Doubling the initial distance the weight is displaced from equilibrium doubles the amplitude, but does not change the period or the phase of the oscillator. So in the general equation of an oscillator:

, only the term changes, and it doubles.

This problem deals with the conservation of energy in the form of a spring. The formula for the conservation of energy is:

We can eliminate initial kinetic energy to get:

We have two forms of potential energy in this problem: gravitational and spring. Therefore, we can rewrite the expression to be:

There is no initial gravitational potential energy and no final spring potential energy. Subsituting our expressions for potential and kinetic energy, we get:

Rearranging for final velocity, we get:

is the original displacement (the same as ).

We have all of the values, so we can solve:

Before the elevator begins to accelerate, the block has a weight of:

While the elevator is accelerating, the block has a weight of:

Now we can say that the weight of the block has effectively increased by 2N.

We can use this increase in force to find out how much the equilibrium position will shift:

Rearranging for x (displacement), we get:

We know that the maximum acceleration of a spring occurs when it is at its maximum displacement. At this point, we can write an expression for the force that the block is under:

Rearranging for the distance, we get:

This is also the point at which the spring is storing its maximum potential energy:

We can substitute our rearranged distance equation into the equation for potential energy:

We also know that the maximum velocity of a spring occurs when it is at the point of equilibrium. Since the spring is on a frictionless surface, we can say that its kinetic energy at this point is equal to the maximum potential energy:

Plugging in the expressions for these variables:

Rearranging for velocity:

We know all of the values, so we can plug and chug:

From the problem statement, we can calculate how much potential energy is initially stored in the spring. Since we are neglecting air resistance, we can say that all of this energy is lost through friction as the system comes to a stop.

Calculating initial potential energy:

Then we can write an expression for the force of friction on the mass:

We can then write an expression of the work done by friction:

Setting this equal to the initial potential energy:

Therefore, the mass travels a total distance of 20 meters before coming to rest. We know that the distance between maximum compression and maximum extension of the spring is 4 meters. Therefore, we can say that the mass travels from one maximum displacement to the other five times, passing through the equilibrium once each time. The mass passes through the equilibrium point 5 times.

We can calculate the initial potential energy of the spring from what we are given in the problem statement:

Assuming the only energy loss of the spring is through internal friction, we can write:

This problem can be solved by the conservation of energy. The block initially has kinetic energy equal to:

When the velocity of the block is zero, all the kinetic energy has been transferred to potential spring energy.

To find the answer to this problem, we'll need to consider the case before the object collides with the spring, and also after it collides. Before the collision, all of the object's energy is in the form of kinetic energy.

Upon collision, the spring will be compressed by the moving object. The maximum displacement, however, will occur the instant the object stops moving. At this point, the object will no longer have any kinetic energy because it has stopped moving. Furthermore, all of the energy has now been transferred into the spring, and is held as spring potential energy.

Also, since we are told in the question stem that no friction is involved, we have a situation in which mechanical energy is conserved. In other words, we can relate the initial and final energies as equal to one another.

Set these two equations equal to each other, and solve for the maximum displacement.

To solve this problem, we must use our knowledge of the relationship between angular frequency, the spring constant, and the mass.

We can then use this, and the derived maximum speed of the object in terms of the amplitude and angular frequency, to find the answer.

Hooke's Law is:

here is in meters and is in newtons.

Solve for to see what the units are:

This means that the units of the spring constant, is:

We can ignore the negative sign when we determine units, as the negative sign only indicates the direction of the force.

To determine which type of sinusoid is used, we have to look at its starting point and where it goes. It starts at its minimum and moves towards its maxima. The simplest model we can use is a negative cosine model, which will require no phase shifts.

At , will be at its minimal value.

However, there is also a counteracting frictional force that decreases the amplitude of the spring over time. One way to do so would be to add an term in front of the cosine, so that we have an ever decreasing amplitude as time moves on.

The only appropriate model is

The amplitude does not change since the student pulls the weight the same displacement. Changing the spring constant changes the frequency of the oscillator. Since the period decreases, the frequency increases, but it does not double. The frequency and the period depend on the square root of the spring constant, so the frequency increases by . In the equation of motion for an oscillator,

, only the frequency , changes.

Since there are two forces acting on the object, and it isn't moving, there is an equality of the two forces. This means that:

Equivalently:

Where is the spring constant, is the displacement of the spring, is the equilibrium position, is mass of the object and is the gravitational constant given as

Since we know that the mass on the spring spring is displaced , the mass is , and we know the gravitational constant we plug in and solve for the spring constant.

Use the equation:

for a mass on a horizontal spring (so no gravity in the direction of motion)

The stretch length will have no effect.

Plug in values:

We know that when the block is at its maximum displacement, it is not moving and therefore has no kinetic energy. Also, we know that the block has its maximum kinetic energy (thus velocity) when it is passing through the springs equilibrium and therefore has no potential energy. Therefore, we can say:

Where the potential energy of a spring is:

And the expression for kinetic energy is:

Substituting in these expressions, we get:

Canceling out the fractions:

No rearranging for mass, we get:

We have all of these values, so we can solve the problem:

The expression for the period of a spring in simple harmonic motion is:

We have the spring constant, so we just need to calculate the mass of the block. For a spring in simple harmonic motion, we know that the maxmium potential energy is equal to the maximum kinetic energy. Therefore we can say:

Rearranging for mass, we get:

Plugging in our values, we get:

Now we can use the expression for period to solve the equation:

Plugging in our values:

We can use the expression for conservation of energy to solve this problem:

We can eliminate initial kinetic energy (block initially at rest) and final potential energy (block at equilibrium of spring) to get:

Substituting in expressions for each of these, we get:

Where initial height is simply the displacement of the spring:

Multiplying both sides of the expression by 2 and rearranging for final velocity, we get:

Plugging in values for each of these variables, we get:

The spring constant is equal to .

Using conservation of energy:

While at the maximum value of stretch (the amplitude), the velocity will be zero. While at the maximum velocity, the stretch will be zero.

Plugging in values:

Solving for

Plugging in values: