A string attached to a fixed wall and a mass on a spring

In summary, the multiple part problem that the expert is stuck on is that they are not sure how to get started with part c or part d. They've worked on part a, but are still stuck on part b. They've managed to get part b, but are not sure how to go forward from there. They believe that part c and part d involve solving a second differential equation, but are not sure how to do so.
  • #1
Old-Alien
6
0
Homework Statement
See attached
Relevant Equations
f(x,t) = A*sin(w*sqrt(u/T))*cos(w*t+phi), etc.
Hi all. Multiple part problem that I'm really stuck on. I'll attach a file.

At first I had attempted the whole problem with the idea that fixed wall was a fixed point, and that the mass on a spring was a "free" point. But I learned later that the mass can't be treated like a "free" point since the spring resists movement, so I'm completely at a loss on how to get started with these problems.

For part a, I tried to readjust certain variables. I knew that sqrt(u/T) is 1/v, and that w/v = k, the wave number. So I rearranged it into Asin(kx)cos(wt+phi), then readjusted cos to sin by making it sin(wt+phi+(pi/2)). I think that's still right but I'm not sure.

For part b, I know that they shared some trig functions to simplify it down some. I did that, then set k to (n+.5)/L to prove that both sides = 0 which is what I thought the boundary condition would be at x = L.

But now I know both of those are wrong, so I'm completely lost and not confident in what to do with c or d...

For part b,
prob3.PNG
 
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  • #2
Hi all. I've worked more on this and have managed to get part a, I think.

Part b I am still stuck.

I've narrowed it down to T * w * sqrt(u/T) * cos(kL) = mw^2 * sin(kL) - Ksp * sin(kL). But I'm not sure how to go forward from here. Can anyone help?
 
  • #3
what do you think the equation they asked you prove represents does it resemble some kind of formula?

edit :

what does
##
m \frac{d^2 \psi(L,t)}{dt^2}
## represent
 
  • #4
timetraveller123 said:
what do you think the equation they asked you prove represents does it resemble some kind of formula?

edit :

what does
##
m \frac{d^2 \psi(L,t)}{dt^2}
## represent
It's a second differential. I had noted that before, and I know that, in springs, F = ma + bv + kx. Here, it's just ma + kx. I also know that Fy in waves = dW/dx, which is the first differential. I could setup an equation that puts them all on the same side. Am I on the right track?
 
  • #5
yes but just forget about f =ma +bv +kx that for now
the only thing you should use here is f=ma so the middle term represent the ma could you show that the other two terms represent force along a suitable axis
 
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  • #6
timetraveller123 said:
yes but just forget about f =ma +bv +kx that for now
the only thing you should use here is f=ma so the middle term represent the ma could you show that the other two terms represent force along a suitable axis
Right. Fy = T*dW/dx, which would be force along the y-axis and force = ma^2. The mass can't move horizontally, so would KspW = 0, thus not playing a factor? Since it is a standing wave, the waves aren't moving left or right, only "up or down," if that makes sense. Sorry, sometimes I am not very good with word choice.
 
  • #7
Old-Alien said:
Fy = T*dW/dx, which would be force along the y-axis and force = ma^2.
yes this correct what do you mean by this
##
ma^2
##
in the y direction there would a component of tension but
Old-Alien said:
The mass can't move horizontally, so would KspW = 0, thus not playing a factor?
the spring does play a factor since it is the y direction
 
  • #8
timetraveller123 said:
yes this correct what do you mean by this
##
ma^2
##
in the y direction there would a component of tension but

the spring does play a factor since it is the y direction
Ah, sorry. By ma^2, I meant that force = m*d^2W/dx^2. By a I just mean the second derivative of W with respect to x.

As for the spring...I suppose then that F = m*d^2W/dx^2 - Ksp*W where I write W at L? When I had it all written down, I had simplified it down to

T*k*cos(kL) = (mw^2 - Ksp)sin(kL)

where k = w*Sqrt(u/T). Do I just need to show that T*k*cos(kL) is equal to the force to prove the boundary condition at x = L is such?
 
  • #9
Old-Alien said:
Ah, sorry. By ma^2, I meant that force = m*d^2W/dx^2. By a I just mean the second derivative of W with respect to x.
ah i see
Old-Alien said:
As for the spring...I suppose then that F = m*d^2W/dx^2 - Ksp*W where I write W at L? When I had it all written down, I had simplified it down to
i think there is some confusion here
##
f = m\frac{d^2 \psi(L,t)}{dt^2}
##
this is f is the force on the mass
now you just need to express this f in terms of physical forces of which there are two here
the vertical component of tension and spring force if equate it with ma you get the equation in b

for part b you just have to work with ##\psi##
you don't have to concern yourself with the actual form of ##\psi##
 
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  • #10
timetraveller123 said:
ah i see

i think there is some confusion here
##
f = m\frac{d^2 \psi(L,t)}{dt^2}
##
this is f is the force on the mass
now you just need to express this f in terms of physical forces of which there are two here
the vertical component of tension and spring force if equate it with ma you get the equation in b

for part b you just have to work with ##\psi##
you don't have to concern yourself with the actual form of ##\psi##
Ok. I think I understand now. Thank you! I have figured out 3c and 3d in the meantime, so I should be good from now on.
 
  • #11
ok and welcome to physics forum
 

1. How does the mass on a spring affect the motion of the string?

The mass on a spring has an impact on the motion of the string as it pulls on the string with its weight, causing the string to stretch and move in response to the force of the mass.

2. What is the relationship between the spring constant and the motion of the string?

The spring constant, which is a measure of the stiffness of the spring, affects the motion of the string. A higher spring constant results in a stiffer spring, which in turn leads to a faster and more intense motion of the string.

3. How does the length of the string affect the motion of the mass on the spring?

The length of the string also plays a role in the motion of the mass on the spring. A longer string will allow for greater displacement and therefore larger oscillations, while a shorter string will restrict the motion of the mass.

4. What factors can affect the frequency of the string's motion?

The frequency of the string's motion is influenced by several factors, including the mass of the object, the stiffness of the spring, and the length of the string. Changes in any of these variables can result in a change in the frequency of the string's oscillations.

5. How does friction impact the motion of the string?

Friction can have a dampening effect on the motion of the string, reducing the amplitude (height) of the oscillations. This is because friction converts some of the energy of the system into heat, causing the string to lose energy and slow down over time.

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