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Alternatively, think of the spring (with spring constant $k$) as two identical elementary springs (each with spring constant $k_0$) in series. It follows that $k=\frac{k_0}{2}$, and hence from the formula for the characteristic frequency, that $f=\frac{f_0}{\sqrt{2}}$. In space without gravity, the midpoint will be at rest due to symmetry, so each side ...
The equation of motion for one mass $m$ attached to a spring with spring constant $k$ (with the other end fixed) on Earth is $$\ddot{x} + \omega^2 x = g$$ where $2 \pi f = \omega = \sqrt{k / m}$. If it's not obvious that the $g$ term on the right hand side doesn't affect the frequency, notice that if we changed our origin so that $z = 0$ when $x = ... 3 i do not understand the part of clamping the center of the coil or cutting it off The solution method appeals to symmetry. Essentially, the system is symmetric (or even about the center of the spring) and this symmetry is exploited to arrive at an answer. If you have taken a first semester in electrostatics, you may be aware of the method of images. ... 1 Yes, alternatively you can use that the potential energy of the spring is transformed into kinetic energy of the object. This is simpler than considering the work$L$done by the spring. The result is however the same: $$U=E_{kin}\\ \Rightarrow \frac{1}{2} k x^2=\frac{1}{2}m v^2\\ \Rightarrow v=x\sqrt\frac{k}{m}$$ 0 Okay My approach was correct and Radius of rotation should be half the length of rod so$2R=l$and solving thereafter will give correct answer. 2 The potential energy is$U\left(r\right) = k \left(r - r_0\right)^2 / 2$, where$r_0$is the equilibrium position of the spring. Since this is a central potential (depends only on$r$), angular momentum$L$is conserved, and we can use the concept of an effective potential energy (see equation 335 here) $$\begin{eqnarray} U_{eff}\left(r\right) &=& ... 2 "the gravitational potential energy should DECREASE as it is converted into kinetic energy, but I guess I'm not really sure why it's negative at position 3." To try and explain this I'll use an analogy. Imagine for a second you have a cat, and you throw the cat up into the air. If you choose to use the ground as a reference point, then the cat starts off ... 0 Assume the point mass to be at position \vec{r}. A spring at \vec{R_i} exerts a force along the line of connection.$$\vec{F_i} = -k_i (\vec{r}-\vec{R_i})$$Summing up and adding a term for gravity, it yields$$ \vec{F} = -\sum_i k_i (\vec{r}-\vec{R_i}) - m\vec{g}$$I can treat the N slinkys (I'm assuming they're parallel) as one large slinky. If ... 0 The direction matters. Consider this diagram: This is using the up is positive convention, so the heights of the objects are positive,$y \gt 0$. A positive force is directed upwards so it accelerates the body upwards while a negative force is directed down so it accelerates the bodies downwards. If you compare this to your diagram the spring is ... 1 You can find the natural frequency of such a system theoretically, if you know the stiffness of the springs. Refer:this example on wolfram. How do you find the stiffness of the springs? Well... do experiments on simple pendulum, find the natural frequency from time period of oscillation, reverse calculate the stiffness of the spring as the mass of the test ... 1 It would depend on how the mass is added to the spring, i.e., you could write for the mass attached to the spring as$M(t)$. Here$M(t)=m_1 + m_2 f(t)$where$f(t)$is some function of time. In principle, you would have to find a realistic$f(t)$based on the dynamical process of how the mass is added, but you could just assume a reasonable function. ... 2 Too long for a comment, so in an answer: Not all the springs are a function of$x_1$and$x_2$, only spring$K_b$is a function of both$x_1$and$x_2$. Spring force is a function of how much a spring is stretched, e.g. how much difference er is between the beginning and the end of a spring, so$x_{begin}$-$x_{end}$or$x_{top}$-$x_{bottom}\$ for this case. ...