# How to find the number of oscillations a block goes through

• kileigh
In summary: After the collision, how much further will it move from equilibrium before reaching maximum compression?
kileigh
Homework Statement
How would I find the number of oscillations that block 2 would undergo if block 2 rested on a patch of ground with friction.
Relevant Equations
Displacement amplitude=x0-n(4f/k)
This is the image provided with the problem, the values given include:
d= 4.00 m, the mass of block one=0.200 kg, speed of block one=8.00 m/s, the period of oscillations for block two without friction=0.140 s, and the spring constant= 1208.5 N/m.

I know how to solve the oscillations if block two was pulled and released:

For this equation, I have the values for k, but I'm not sure how to change it so rather than being from the distance pulled(x0), instead of for the energy from block one when they collide.
Thanks!

So block one, having mass 2 kg and speed 8 m/s, so kinetic energy 65 Joules, hits block 2, compressing the spring. Are we not told the mass of block 2?

I calculated the mass of block two with the period without friction. I got a mass of 0.6 kg, but I was not sure if that would change since the period of the block is technically changing when friction is considered.

Well, the mass will not change just because there is friction! My first thought was to use the period and mass to calculate k buy I didn't notice that we are given k but not the mass!

kileigh said:
I calculated the mass of block two with the period without friction. I got a mass of 0.6 kg, but I was not sure if that would change since the period of the block is technically changing when friction is considered.
You don't need to know the period with friction.

There must be rather more to the whole question. If you were to provide the whole we would also be able to confirm that you've not overlooked anything, like further interaction with block 1.

haruspex said:
You don't need to know the period with friction.

There must be rather more to the whole question. If you were to provide the whole we would also be able to confirm that you've not overlooked anything, like further interaction with block 1.
This was the whole first question, but my professor asked the question in my orginal post: Figure 15.15 shows block 1 of mass0.200 kg0.200 kg sliding to the right over a frictionless elevated surface at a speed of8.00 m/s8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant1208.5 N/m1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of0.140 s0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distancedd from the base of that surface after falling heighth=4.90 mh=4.90 m. What is the value ofdd?

kileigh said:
This was the whole first question, but my professor asked the question in my orginal post: Figure 15.15 shows block 1 of mass0.200 kg0.200 kg sliding to the right over a frictionless elevated surface at a speed of8.00 m/s8.00 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant1208.5 N/m1208.5 N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of0.140 s0.140 s, and block 1 slides off the opposite end of the elevated surface, landing a distancedd from the base of that surface after falling heighth=4.90 mh=4.90 m. What is the value ofdd?
Ok, so there will be no further interaction with block 1.
As I posted, you don't need the period with friction. You found the mass from the period without friction, and you can use that in your formula for the number of oscillations.
I assume you found the initial speed of block 2 ok.

But which equation should I use, my original equation asks for the distance the spring is pulled and I'm not sure what to put for the x0.

kileigh said:
But which equation should I use, my original equation asks for the distance the spring is pulled and I'm not sure what to put for the x0.
During one half oscillation the frictional force is constant. This means it is like vertical oscillation under gravity, so it is SHM. Find the SHM equation for that and use the known initial speed. But note that the initial position will not be the equilibrium position.

Would it be similar to my original equation? Sorry I struggle with finding my SHM equations.

kileigh said:
Would it be similar to my original equation? Sorry I struggle with finding my SHM equations.
As I posted, for the purposes of the left to right movement, we can treat friction as a constant force to the left. Where would the equilibrium position be on that basis?
After the collision, how much further will it move from equilibrium before reaching maximum compression?

## 1. How do I measure the number of oscillations a block goes through?

To find the number of oscillations, you can use a stopwatch and count the number of times the block moves back and forth in a set amount of time. Alternatively, you can record the movement using a video camera and use slow-motion playback to count the oscillations.

## 2. What is the formula for calculating the number of oscillations?

The number of oscillations can be calculated using the formula: N = t/T, where N is the number of oscillations, t is the total time elapsed, and T is the period of oscillation.

## 3. Can the number of oscillations change over time?

Yes, the number of oscillations can change over time if the amplitude or frequency of the oscillations changes. For example, if the amplitude decreases, the number of oscillations will decrease as well.

## 4. How does the mass of the block affect the number of oscillations?

The mass of the block does not affect the number of oscillations. The number of oscillations is only affected by the amplitude and frequency of the oscillations.

## 5. Is there a difference between the number of oscillations and the number of cycles?

No, the number of oscillations and the number of cycles are the same. Both terms refer to the number of times the block completes a full back-and-forth motion.

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