Travel time of a particle suspended from an elastic string

In summary, the author calculates the time it takes for a mass to rise from 1m below the string to the string's equilibrium height. They make an error in their calculation and need to be corrected.
  • #1
gnits
137
46
Homework Statement
To find the time of travel of a particle on an elastic string
Relevant Equations
F=ma
Hi,

Can anyone please help me with the following:

Q3.png


I have found the velocity of projection, no problem, it is v = 2*sqrt(10)

Also, in obtaining this value, I have also found the extension in the string when in equilibrium, it is x = 2

Now on to the time of flight.

The given answer is: t = ( PI/(2 * sqrt(40) ) + sqrt(2)/sqrt(10)

So the motion is in two parts, one when the string is stretched and so pulling on the mass, and one where the string is unstretched and so the mass is simply a projectile.

I am going to try to calculate the time taken to rise from 1m below A to A. This is the part of the motion which is that of a projectile. And I believe form the given answer that I zshould expect to obtain sqrt(2)/sqrt(10). I do not.

Hopefully, someone can show me where this method is in error:

Here's a diagram:

Diag.png


On the left is just the string, on the right is the string stretched by the 2g weight.

Let the general position of the particle be y below the unstretched level. The tension in the string is T. Hookes law gives T = Kx/a where K is the modulus of elasticity (here = 10), a is the natural length (here = 1), and the extention is x (in the diagram this is equal to y).

So we have T = 10y

Now equating forces gives (F = ma):

10y - 2g = 2 * acceleration

And so we have: acceleration = 5y - g = 5y - 10

Now, as accerleration = dv/dt = (ds/dt)(dv/ds)=v (dv/ds) we have:

v dv/dy = 5y - 10 and so separating variables gives:

integral(v dv) = integral(5y-10 dy) and so we have:

(1/2)v^2 = (5/2)y^2 - 10y + c

And using v = 2*sqrt(10) when y = 2 leads to c = 30

and so we have (1/2)v^2 = (5/2)y^2 - 10y + 30 which is

v^2 = 5y^2 - 20y + 60

So we can now put in y = 0 to obtain the speed of the mass at the moment that the string becomes slack. From here I use v = u - gt to obtain t - the time to go from slack to A. This leads to a different answer from the book.

Thanks for any help in pointing out my error,
Mitch.
 
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  • #2
gnits said:
the velocity of projection, no problem, it is v = 2*sqrt(10)
Not so. The string helps it up.

Edit2: having realized the 2g is a typo for 2kg, I withdraw the above.

Edit: also, I would take the 1m as the relaxed length. If you provide the book answer we can determine which reading is correct.
 
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  • #3
gnits said:
Now, as accerleration = dv/dt = (ds/dt)(dv/ds)=v (dv/ds) we have:

v dv/dy = 5y - 10

Be careful with signs here. In setting up F = ma you chose the upward direction as positive. But y increases as you move downward.

You may write dv/dt = (dv/dy)(dy/dt), as this is just the chain rule. But does dy/dt = v or does dy/dt = -v?
 
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  • #4
TSny said:
Be careful with signs here. In setting up F = ma you chose the upward direction as positive. But y increases as you move downward.

You may write dv/dt = (dv/dy)(dy/dt), as this is just the chain rule. But does dy/dt = v or does dy/dt = -v?

Thank you very much, that was it, the writing of dy/dt as v was the mistake. I should have written it, as you say, as -v. Then I obtain the book answer of sqrt(2)/sqrt(10). Thanks so much for spotting this. I will be more careful in future.

My remaining task is to find the time taken to go from the initial point of projection to the point 1m below A. The form of the answer PI/(2 * sqrt(20)) makes me think that this part of the motion should be treated as Simple Harmonic Motion. Is this in fact the case? If the particle had simply been released then I can see that it would be, but is it still Simple Harmonic Motion if the particle is initially projected with some velocity?

(If I were to try to use my equation of "v as a function of y" to solve for "v as a function of t" I would get: dy/dt=sqrt(20y-5y^2+20) and this leads to (20y - 5^2+10)^(-1/2) dy = dt and integrating the LHS isn't pretty and given that the context of the question is a chapter on SHM I'm thinking that there must be a smarter way)

Thanks for any help.
 
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  • #5
gnits said:
Thank you very much, that was it, the writing of dy/dt as v was the mistake. I should have written it, as you say, as -v. Then I obtain the book answer of sqrt(2)/sqrt(10). Thanks so much for spotting this. I will be more careful in future.

My remaining task is to find the time taken to go from the initial point of projection to the point 1m below A. The form of the answer PI/(2 * sqrt(20)) makes me think that this part of the motion should be treated as Simple Harmonic Motion. Is this in fact the case? If the particle had simply been released then I can see that it would be, but is it still Simple Harmonic Motion if the particle is initially projected with some velocity?

(If I were to try to use my equation of "v as a function of y" to solve for "v as a function of t" I would get: dy/dt=sqrt(20y-5y^2+20) and this leads to (20y - 5^2+10)^(-1/2) dy = dt and integrating the LHS isn't pretty and given that the context of the question is a chapter on SHM I'm thinking that there must be a smarter way)

Thanks for any help.

Ok, I was not too far off, my mistake was thinking that the integral was unpleasant, in fact the integral is simply -(1/sqrt(5))*asin((2-x)/(2*sqrt(2))) and it all works out to the book answer.

Thanks again for all of your help,
Mitch.
 
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  • #6
I love this. The youngster does and shows his work, is well assisted, and is appreciative. This is a big part of why PF is a great site.
 
  • #7
The mass is given as 2g, the relaxed length as 1m and the modulus as 10N. In equilibrium the extension should be 2mm, not 2m.
I presume the 2g was a typo for 2kg.
 

Related to Travel time of a particle suspended from an elastic string

1. How does the length of the elastic string affect the travel time of the suspended particle?

The longer the elastic string, the longer the travel time of the suspended particle. This is because a longer string allows for more distance for the particle to travel before being pulled back by the elastic force.

2. Does the mass of the suspended particle impact its travel time?

Yes, the mass of the suspended particle can affect its travel time. A heavier particle will require more force to be pulled back by the elastic string, thus resulting in a longer travel time compared to a lighter particle.

3. How does the elasticity of the string influence the travel time of the particle?

The more elastic the string is, the shorter the travel time of the suspended particle. This is because a more elastic string can exert a greater force to pull the particle back, resulting in a faster overall travel time.

4. Can the direction of the particle's travel affect its travel time?

Yes, the direction of the particle's travel can impact its travel time. A particle traveling in a straight line will have a shorter travel time compared to a particle traveling in a curved or zigzag path, as the latter requires more energy and time to change direction.

5. How can air resistance affect the travel time of the suspended particle?

Air resistance can act as a drag force on the suspended particle, slowing it down and thus increasing its travel time. This effect is more significant for larger particles or particles traveling at higher speeds.

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