Calculate the spring constant of oscillating mass on a spring colliding with a wall

[Aurelius120]

Homework Statement

Block of mass 0.9kg. Spring(k) on frictionless floor. Spring compressed to $\sqrt 2$cm. Block at $1/\sqrt 2$cm from wall. Collision is elastic and time period of motion 0.2s. Approximate value of k?

Relevant Equations

$x=A sin(\omega t+\phi)$
$\omega = \sqrt{\frac{k}{m}}$

So I proceed as:
Total time for 1 oscillation is 0.2s
$$\frac{1}{\sqrt{2}}=\sqrt{2} \sin ({\omega t_1})$$
$$\sqrt{2}=\sqrt{2} \sin ({\omega t_2})$$
Therefore
$$\omega t_2=\frac{\pi}{2}$$
$$\omega t_1=\frac{\pi}{6}$$
$$\omega ×2(t_1+t_2)=2×\left( \frac{\pi}{2}+\frac{\pi}{6}\right) $$
Since $(t_1+t_2)$ is time to complete half oscillation, $2(t_1+t_2)=0.2$
$$\implies \omega^2=\frac{k}{m}=\frac{k}{0.9}=\left( \frac{4\pi}{3×0.2}\right)^2$$
Therefore $k=400N/cm$

The answer given is $100Nm^{-1}$

How to obtain the correct answer?

 

[PeroK]

Did you not notice in the question that the mass is bouncing off the wall? It's not only oscillating freely!

 

[Aurelius120]

Yes but that time is not mentioned. So I assumed the ball velocity was simply reversed in direction at the point without any other change. Therefore it was an SHM but took $2(t_2-t_1)$ less than original of $4t_2$ Are you saying 0.2 seconds is the time of contact for collision?

 

[PeroK]

Yes but that time is not mentioned neither is the coefficient of restitution. So I assumed the ball velocity was simply reversed in direction at the point without any other change. Are you saying 0.2 seconds is the time of contact for collision?

What answer would you get if you ignored the collision with the wall altogether?

 

[pasmith]

I don't think you can have periodic motion unless the collision with the wall is perfectly elastic, so that no energy is lost; the velocity then simply changes sign when the block collides with the wall.

 

[scottdave]

What if the block just barely touches the wall opposite the spring?

 

[Aurelius120]

I don't think you can have periodic motion unless the collision with the wall is perfectly elastic, so that no energy is lost; the velocity then simply changes sign when the block collides with the wall.

Yes the question does say all collisions are elastic. See the image for full question

What if the block just barely touches the wall opposite the spring?

The amplitude of oscillation ie compression in spring is $\sqrt 2$ and distance between wall and block is $1/\sqrt 2$

 

[Aurelius120]

What answer would you get if you ignored the collision with the wall altogether?

You mean if there was no collision and just SHM of time period 0.2s $$(0.2)^2=\frac{4\pi^2×0.9}{k}$$ which is not $100 Nm^{-1}$

 

[PeroK]

You mean if there was no collision and just SHM of time period 0.2s $$(0.2)^2=\frac{4\pi^2×0.9}{k}$$ which is not $100 Nm^{-1}$

So, how did you factor in the collision with the wall? Your OP is just equations with no explanation of what you're doing.

 

[PeroK]

Homework Statement: Block of mass 0.9kg. Spring(k) on frictionless floor. Spring compressed to $\sqrt 2$cm. Block at $1/\sqrt 2$cm from wall. Collision is elastic and time period of motion 0.2s. Approximate value of k?
Relevant Equations: $x=A sin(\omega t+\phi)$
$\omega = \sqrt{\frac{k}{m}}$

View attachment 336298
So I proceed as:
Total time for 1 oscillation is 0.2s
$$\frac{1}{\sqrt{2}}=\sqrt{2} \sin ({\omega t_1})$$
$$\sqrt{2}=\sqrt{2} \sin ({\omega t_2})$$
Therefore
$$\omega t_2=\frac{\pi}{2}$$
$$\omega t_1=\frac{\pi}{6}$$

I can't figure out what you're doing here at all, I'm sorry to say.

 

[Aurelius120]

So, how did you factor in the collision with the wall? Your OP is just equations with no explanation of what you're doing.

Sorry. I will add them.

Homework Statement: Block of mass 0.9kg. Spring(k) on frictionless floor. Spring compressed to $\sqrt 2$cm. Block at $1/\sqrt 2$cm from wall. Collision is elastic and time period of motion 0.2s. Approximate value of k?
Relevant Equations: $x=A sin(\omega t+\phi)$
$\omega = \sqrt{\frac{k}{m}}$

View attachment 336298

So I proceed as:

Since velocity is reversed at the wall. I assumed the SHM to be complete half oscillation with time $2t_2$(to and fro) on the spring-ward side where it completes displacement of $A(=\sqrt 2)$

The SHM is incomplete on the other side as the wall reverses the direction of motion of block with an elastic collision and a total time of $2t_1$ is taken coming and going to the wall at $\frac{1}{\sqrt{2}}$. To and fro times are equal since velocity is reversed and acceleration & displace from mean position unchanged

I assumed negligible contact time of collision

Total time for 1 oscillation is 0.2s
$$\frac{1}{\sqrt{2}}=\sqrt{2} \sin ({\omega t_1})$$
$$\sqrt{2}=\sqrt{2} \sin ({\omega t_2})$$
Therefore
$$\omega t_2=\frac{\pi}{2}$$
$$\omega t_1=\frac{\pi}{6}$$
$$\omega ×2(t_1+t_2)=2×\left( \frac{\pi}{2}+\frac{\pi}{6}\right) $$
Since $(t_1+t_2)$ is time to complete half oscillation, $2(t_1+t_2)=0.2$
$$\implies \omega^2=\frac{k}{m}=\frac{k}{0.9}=\left( \frac{4\pi}{3×0.2}\right)^2$$
Therefore $k=400N/cm$

The answer given is $100Nm^{-1}$

How to obtain the correct answer?

 

[PeroK]

Okay, but you have $x = 0$ at $t = 0$. Which is not right. At $t =0$, the mass is at the maximum displacement.

 

[Aurelius120]

Okay, but you have $x = 0$ at $t = 0$. Which is not right. At $t =0$, the mass is at the maximum displacement.

Phase difference might affect equation of SHM but I don't think it will change the value of time required to move from $-A$ to $0$ or $0$ to $A/2$.
These are the only two solutions I found on the web but I can't make heads or tails of these:
Solution 1
Thats $k=100\pi^2/9$ btw

Solution 2

Does it mean the book might have given a wrong answer or I am misinterpreting the question ?

 

[PeroK]

The first solution is what I did. The second solution looks like nonsense to me.

 

[kuruman]

The first solution is what I did.

Me too. The key here is to realize that the distance from the initial position of the mass to the wall is $\frac{3}{2}A$ where $A$ is the amplitude and then find the time required to cover that distance as a fraction of the unobstructed period of oscillations.

 

[haruspex]

Phase difference might affect equation of SHM but I don't think it will change the value of time required to move from $-A$ to $0$ or $0$ to $A/2$.
These are the only two solutions I found on the web but I can't make heads or tails of these:
Solution 1View attachment 336338
Thats $k=100\pi^2/9$ btw

Solution 2
View attachment 336339

Does it mean the book might have given a wrong answer or I am misinterpreting the question ?

Why do you think this means the book's answer might be wrong? Both answers agree with the book.
Do you understand the first solution?
The second solution is unintelligible because it seems to use T to mean two different things.

Since velocity is reversed at the wall. I assumed the SHM to be complete half oscillation with time $2t_2$(to and fro) on the spring-ward side where it completes displacement of $A(=\sqrt 2)$

Not sure what you are saying there. The distance from the start position to the wall is 1/4 of a full unobstructed oscillation.

 

[Aurelius120]

Why do you think this means the book's answer might be wrong? Both answers agree with the book.

I couldn't find it after considerable effort and couldn't understand the solution. Plus there was yet another solution on the internet that agree with my incorrect solution.

Not sure what you are saying there. The distance from the start position to the wall is 1/4 of a full unobstructed oscillation

What I meant was that $2t_2$ is the time for $-A\rightarrow 0$ and later from $0\rightarrow -A$ and $2t_1$ is time for $0 \rightarrow +A/2 \rightarrow 0$

Therefore $2(t_1+t_2)$ is time for one full oscillation which is given as $0.2s$ in the question(or so I think).

 

[Aurelius120]

Do you understand the first solution?

Nope.

 

[PeroK]

I couldn't find it after considerable effort and couldn't understand the solution. Plus there was yet another solution on the internet that agree with my incorrect solution.

What I meant was that $2t_2$ is the time for $-A\rightarrow 0$ and later from $0\rightarrow -A$ and $2t_1$ is time for $0 \rightarrow +A/2 \rightarrow 0$

Therefore $2(t_1+t_2)$ is time for one full oscillation which is given as $0.2s$ in the question(or so I think).

Here's an equivalent problem. The spring is compressed by $\sqrt 2$cm and the mass is released. The time taken to move an initial distance of $\frac 1 {\sqrt 2}$ cm from the point of release is measured as $0.1$s. What is the spring constant?

 

[haruspex]

What I meant was that $2t_2$ is the time for $-A\rightarrow 0$ and later from $0\rightarrow -A$ and $2t_1$ is time for $0 \rightarrow +A/2 \rightarrow 0$

Therefore $2(t_1+t_2)$ is time for one full oscillation which is given as $0.2s$ in the question(or so I think).

I assume you are defining A as the initial compression distance, but I do not know how you are defining the zero position. If we take 0 as the relaxed position of the spring, the mass executes $-A\rightarrow -A/2$ and back for one complete cycle.

 

[PeroK]

Nope.

If this problem is too difficult for you, you need to find some simpler problems. Not understanding a relatively well explained solution is a sign, IMO, that the problem is too hard for the time being.

That said, the concept of the mass bouncing elastically off a wall seems a bit far fetched to me (although, at least it wasn't bouncing off an elephant!). Maybe that confused you.

In any case, the fundamental problem was using $A\sin(\omega t)$, which is only valid where the mass is at the midpoint at time $t = 0$. If the mass is at maximum displacement at $t = 0$, then you must use $\pm A\cos(\omega t)$ for the displacement.

 

[Aurelius120]

I assume you are defining A as the initial compression distance, but I do not know how you are defining the zero position. If we take 0 as the relaxed position of the spring, the mass executes $-A\rightarrow -A/2$ and back for one complete cycle.

You mean $+A/2$. Also yes 0 is the relaxed position ('would be mean without wall' position)
$t_2$ is time from $-A\rightarrow 0\ (\&\ vice-versa)$ and $t_1$ is from $0 \rightarrow +A/2 \ (\& \ vice-versa)$
I realise my variables weren't mnemonic.

Therefore $(t_1+t_2)$ is time for $-A\rightarrow 0 \rightarrow +A/2$

 

[Aurelius120]

If this problem is too difficult for you, you need to find some simpler problems. Not understanding a relatively well explained solution is a sign, IMO, that the problem is too hard for the time being.

That said, the concept of the mass bouncing elastically off a wall seems a bit far fetched to me (although, at least it wasn't bouncing off an elephant!). Maybe that confused you.

In any case, the fundamental problem was using $A\sin(\omega t)$, which is only valid where the mass is at the midpoint at time $t = 0$. If the mass is at maximum displacement at $t = 0$, then you must use $\pm A\cos(\omega t)$ for the displacement.

Back to basics then
Thanks

 

[PeroK]

Back to basics then
Thanks

Try my alternative problem:

Here's an equivalent problem. The spring is compressed by $\sqrt 2$cm and the mass is released. The time taken to move an initial distance of $\frac 1 {\sqrt 2}$ cm from the point of release is measured as $0.1$s. What is the spring constant?

 

[haruspex]

You mean $+A/2$.

No, I don't. If the starting position is -A then the wall is at -A/2. 0 is to the right of the wall.

 

[PeroK]

No, I don't. If the starting position is -A then the wall is at -A/2. 0 is to the right of the wall.

In solution 1) above, the positive x-direction is to the left. In fact, I did that as well. It just seemed simpler.

 

[haruspex]

In solution 1) above, the positive x-direction is to the left. In fact, I did that as well. It just seemed simpler.

Yes, but I was responding to post #17, which uses positive right.

 

[Aurelius120]

Try my alternative problem

$-A+A/2=-A\cos (\omega t)$
$$-\sqrt{2}-\frac{-1}{\sqrt 2}=-\sqrt{2}\cos (\omega t)$$
$$\implies 1-\frac{1}{ 2}=\cos (\omega t)\implies \omega (0.1)= \frac{\pi}{3}$$
$$\implies \frac{k}{m}=\frac{k}{0.9}=\frac{\pi^2}{(0.3)^2} \implies k=10\pi^2$$

No, I don't. If the starting position is -A then the wall is at -A/2. 0 is to the right of the wall.

So diagram represents the compressed spring at $-A$ about to be released and the wall at $-A/2$ and therefore @PeroK 's analogue of the problem. I am getting it to some extent now.My interpretation was that the given diagram was of the relaxed spring (at $0$) which would be compressed to left $(-A)$ and released. The wall being on the right $(+A/2)$. That's why I got all that mess using $t_1$ , $t_2$ and what not.

 

[PeroK]

$-A+A/2=-A\cos (\omega t)$
$$-\sqrt{2}-\frac{-1}{\sqrt 2}=-\sqrt{2}\cos (\omega t)$$
$$\implies 1-\frac{1}{ 2}=\cos (\omega t)\implies \omega (0.1)= \frac{\pi}{3}$$
$$\implies \frac{k}{m}=\frac{k}{0.9}=\frac{\pi^2}{(0.3)^2} \implies k=10\pi^2$$

That's good. That's the first aspect of the original problem.

The second aspect of the original problem was to recognise that the motion after the collision was the reverse of the initial motion. Then, you can justify that $t_1 = t_2 = 0.1s$. And, you can do the calculation you just did.

There are several ways, I guess, to see that the elastic collision with the wall simply reduces the motion to the initial and final stages of what would have been the full SHM.

Solution 1) above doesn't give a justification, as far as I can see. What would be your explanation?

 

[Aurelius120]

Solution 1) above doesn't give a justification, as far as I can see. What would be your explanation?

Diagram represents the compressed spring at −A about to be released and the wall is at −A/2 and therefore @PeroK 's analogue of the problem. I am getting it to some extent now.

 

[Aurelius120]

But a rather curious coincidence:
If I go by my first interpretation:

My interpretation was that the given diagram was of the relaxed spring (at $0$) which would be compressed to left $(-A)$ and released. The wall being on the right $(+A/2)$. That means one oscillation will be $$(-A)\rightarrow 0\rightarrow \frac{+A}{2} \rightarrow (-A)$$

Period of motion =0.2s given in question is time for $(-A)\rightarrow 0 \rightarrow \frac{+A}{2} \rightarrow (-A)$
Then $$-A\cos (\omega t)=\frac{+A}{2}$$
$$\implies -\sqrt 2 \cos (\omega t)=\frac{+1}{\sqrt 2}$$
$$\implies \cos (\omega t)=\frac{-1}{2} \implies \omega t=\frac{2\pi}{3}$$
According to question time period of oscillation, 2t=0.2s therefore $$2\omega t= 0.2\omega = \frac{4\pi}{3}$$
This is the same incorrect answer I obtained in POST(1,11,22) and therefore this method is equivalent to that.

Period of motion =0.2s given in question is time for$(-A)\rightarrow 0 \rightarrow \frac{+A}{2}$
Here I take t = 0.2s

Then, $$\omega t=0.2 \omega=\frac{2\pi}{3}\implies \frac{k}{0.9}=\frac{4\pi^2}{0.6^2}$$
Now this gives the correct answer.

This could either be a coincidence or even my initial interpretation is correct if period of motion in the question refers to time taken before first collision. [(-A) to (+A/2)]Could this question have two meanings? (POST 28)

 

[haruspex]

before first collision. [(-A) to (+A/2)]

For the third time, it is not [(-A) to (+A/2)].
"Spring compressed to √2cm. "
A = √2cm.
"Block at 1/√2cm from wall."
Block starts at A/2 from wall; equilibrium is A from starting position, so A/2 to the right of the wall. The motion is an oscillation between -A and -A/2.