# Tag Info

11

The pulleys make it trickey. You can see how this works by looking at the lengths of the individual rope pieces. When the cart moves 1 meter to the right the bottom part gets 1 meter shorter. The top part also gets 1 meter shorter. Since the rope can't change length the left part (that is attached to $m$) gets longer by 2 meters. So for every meter that the ...

3

I believe your answer is correct. I assume there is a mistake in the textbook, and they meant to write that $h=4R$. Then the textbook answer would have been right.

2

Are you talking about the Catenary shape? The general equation is $$y(x) = a \left( \cosh \left( \tfrac{x}{a} \right) -1 \right)$$ where $y(0)=0$ is the lowest point on the curve, and the parameter $a$ defines how much it bends. What remains constant along the curve is the horizontal component of tension $H$, which can be used to find $a$ $$a = \frac{H}{w}... 2 There is one Lagrangian for both x's, but there is a separate Euler-Lagrange equation for each of x_1, x_2. 1 Take a look at the picture bellow (adapted from here https://guitar-auctions.co.uk/wp-content/uploads/2018/02/lot0126.jpg) where I give you the recursive idea to calculate the length x_{n+1} from x_n for the frets on a guitar. x_0 is simply the length of your string (in your case x_0 = L = 1  m). The density of the string does not matter. Why? As you ... 1 The frequency spectrum is given by:$$f_n=\frac{n}{2L}\sqrt{\frac{T}{\rho}}$$and:$$f_1=\frac{1}{2L}\sqrt{\frac{T}{\rho}}$$where T is the string tension. Alternatively, write:$$f_n=\frac{1}{2x(n)}\sqrt{\frac{T}{\rho}}$$So that:$$\boxed{\frac{f_n}{f_1}=\frac{x(n)}{L}}$$Calculate the x(n) from there. 1 The catenary for equal-height pinned ends satisfies$$s=a\sinh\left(\frac{L}{a}\right),$$where s is the length of the rope, L is the distance between the ends, and$$a=\frac{T_\text{horizontal}}{\rho g},$$where T_\mathrm{horizontal} is the horizontal force, \rho is the rope density per unit length, and g is the gravitational acceleration. This is ... 1 I am pretty sure you are correct. I tried to find the mechanical advantage by using a FBD and got it to be 4:1. But just to be sure I poked around the internet a bit and found this great site. The MA is given in section 6-8.3 . 1 The proof can be found in any elementary text on waves. You can do it as follows: We follow the notation as in the figure. Net force upward is$$F_x(t)=T_2\sin\theta_2-T_1\sin\theta_2 We want to express $F_x(t)$ in terms of $\psi(z,t)$ and its space derivative $\partial \psi/\partial z$. Now we make an approximation here, In the small-oscillation ...

1

Your first equation is not particularly obvious to me, but if you take the distances from where the cord leaves the pulley to where it connects to the mass: vertical, y, horizontal, x, and length, L, then: y = L cos(θ) and x = L sin(θ). Take the derivative of each equation with respect to, t, and eliminate the dθ/dt, then you get dy/dt = (dL/dt)/cos(θ).

1

Generally, when it comes to constraint relation of inextensible strings the components of velocity ,acceleration etc,of a particle attached to it are taken along the direction of the string as its length cannot change.It is wrong to take the component of the veleocity of the string in the direction of motion of the object.Furthermore, your equation would ...

1

If we assume Hooke's law holds, then for a spring constant $k$ with resting length $\ell$, the spring force is given by $F=\pm k(x_R-x_L-\ell)$, where $x_R$ and $x_L$ are the positions of the right and left ends of the spring respectively (the $\pm$ sign is to take care of which side of the spring you are looking at). Now, if you assume identical masses $m$ ...

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