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

[Old-Alien]

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,

 

[Old-Alien]

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?

 

[timetraveller123]

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

 

[Old-Alien]

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?

 

[timetraveller123]

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

 

[Old-Alien]

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.

 

[timetraveller123]

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

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

 

[Old-Alien]

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?

 

[timetraveller123]

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

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 $\psi$

 

[Old-Alien]

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.

 

[timetraveller123]

ok and welcome to physics forum