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Basically this is just the solution to the recurrence formula, look here. They show how to solve recurrence relation using difference equation.

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When $\theta = 0$ the spring extension is zero and take this to be the zero of gravitational potential energy. With the rod at an angle $\theta$ from the geometry of the system find the extension $x$ of the spring and the vertical height $h$ through which the centre of mass of the rod has fallen. The potential energy of the system $V = \frac 12 k x^2 - ... 0 My guess:$\theta = 0$might not be equilibrium cause if you stop holding the A box, it will fall. at$\theta = 90$might be neutral equilibium (I assume the spring can pass through the wheel to the side of box A), since there is no tension in the spring, and the normal force on the B box balances the forces of gravity(assuming the B box is restricted to ... 0 A force$F$compresses the springs by an amount$X$where$F = (k_1+k_2)X$and the same force compresses spring$1$by an amount$x$where$F = k_1 x$. 2 Find the center of mass (CM) of the two connected particles. Then, determine the distance (r) from the center of mass of the part of the spring where the third particle hit it. Then, if you could allow a collision time and force while in contact, and assume the collision to be frictionless (this is might be hard for point particles and thin spring, cause ... 4 The way I understand the setup: The block is initially at rest at some height above the spring. The spring is initially at rest oriented vertically with one end on the ground and at it's natural length (though if its not a massless spring it would compress somewhat due to its own weight). Then, the block is dropped, it lands on the spring compressing it ... 1 Let's look at what happens right when the falling jumper passes that equilibrium point (EP). At that point, as you correctly pointed out, the force on them (up, from the bungee) is equal to the force down (gravity). So, there is a total net force of zero. However, they already have momentum from falling. Newton's first law tells us that if there's no net ... 0 The work done onto the spring is$dE/dt=F(t) \dot x(t)$. You should not look at the direction of$F$alone, but at the the direction of the motion as well. 1 Now the real problem I'm having is trying to decide what the forces are acting on the system in order to come up with my differential equation? As there are no external forces the linear momentum of the system will be constant of motion and the constant can be taken as zero as well. If m1 and m2 are the masses at any time described by position ... 1 To do the equations of motion you need positions from an inertial reference frame. Say a wall far away. Call positions of the two masses$x_1$and$x_2. The spring force (tension is positive) is $$F = k (x_2-x_1 )$$ and the two equations of motion \begin{align} F & = m_1 \ddot{x}_1 \\ -F & = m_2 \ddot{x}_2 \end{align} All this is trivial. ... 1 Assuming the spring is "ideal" (massless) you actually have 2 masses. You can describe your problem as the motion of the center of mass, and either of the masses. And if no external force is exerted on your system, you are only left with the motion of 1 mass relative to the center of the mass of the system. Let's say your masses are m1, m2, and the spring ... 0 I think the book answer is correct. The small angle approximation is not necessary : sin\theta $does not appear in the analysis.$ x = r\theta $is exact. The motion is a rotation about the point of contact with the horizontal plane. Using the Parallel Axes Theorem, the moment of inertia about this point is$\frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$. ... 3 The equation for the period$T$comes by using Newton's second law$F=ma$to obtain the equation of motion of the spring-mass system $$-kx =ma \Rightarrow a=-\frac k m$$ where$x$is the displacement from a fixed point and$a$is the acceleration. This equation is of the form$a=-\omega^2 x$where$\omega$is a constant of the simple harmonic motion. It ... 6 In this equation$\omega$does not refer to the speed of angular motion, but the frequency of oscillation when measured in angular terms (usually radians/sec, but it can be degrees/sec). Frequency is usually measured in cycles per second (Hertz), but it is sometimes more conveniently measured in angular terms, when it is called angular frequency. The angle ... 5 Imagine a point$P$moving on a circle of radius$R$with angular velocity$\omega$. The projection of$P$onto the$y$-axis is: $$y=R\sin \theta=R\sin \omega t$$ The point$P'$is in simple harmonic oscillation (SHO). For the spring mass system it just so happens that: $$x=A\sin \omega t$$ where: $$\omega=\sqrt{\frac{k}{m}}$$ So although there is no ... 2 The book answer looks wrong to me. I think$k$can only appear in combination with$x$, since the tension$T=kx$in the spring is relevant here but$x$on its own is not. It could be relevant as a geometrical factor, but there is no length marked$x$in the diagram. I agree with your answer - except that you should have$k$in there with$x$. I do not ... 0 The discussion which has ensued from the question here and in A conceptual doubt regarding forced oscillations and resonance hinges on how resonance is defined for particular situations and what is meant by the natural frequency of the driven system. An often used definition of resonance is: Resonance is the maximum steady state response of a driven ... 0 Use below free body diagrams and note that$P_2=\rho gh_2$and$h_2=h+y-xand y is lower piston displacement. Note that volume of the displaced water for both pistons is equal. Explanation of the answer: At first, let’s review the problem description. We have a container is shown in the problem description above. There is water inside the container. The ... 2 You can take the two equations of motion \begin{align} k (x_2 - x_1) = m_1 \ddot{x}_1 \\ k (x_1-x_2) = m_2 \ddot{x}_2 \end{align} and transform them using their centroid location, and distance \left. \begin{align} x_c & = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}\\ x_d & = x_2-x_1 \end{align} \right\} \begin{aligned} x_1 &= x_c - \frac{m_2 ... -1 There is only one degree of freedom (the distance between the masses) not two, and therefore only one natural frequency (given by your 1st eqn). Since there are no external forces, the CM of the system does not move. (Sorry but I do not see what all the fuss is about here. Why make such a simple problem so complicated?) 2 Suppose I have two masses m1, m2 connected by one spring of stiffness k The Lagrangian of the system isL = \frac{1}{2}m_1\dot q_1^2 + \frac{1}{2}m_2\dot q_2^2 - \frac{1}{2}k(q_1 - q_2)^2$where$q_1$and$q_2$are the coordinates of$m_1$and$m_2$respectively. Now, consider a change of coordinates to the normal coordinates$Q_1$and$Q_2\$ ...

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The other natural frequency is indeed zero! Natural frequencies of zero corresponds to vibrational modes of rigid body motion. Rigid body motion is not a vibrational motion in itself, but still arises in the modal analysis of certain systems such as the one above. The reason you get a rigid body mode is because you are able to move the system as a whole, and ...

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It is difficult to give a definitive answer. The spring constant refers to an ideal spring which obeys Hooke's Law (extension proportional to load). Real springs and especially elastic materials such as bungee cord do not necessarily follow this law. Usually they do initially for 'smallish' loads, but they depart from it gradually at large loads. What ...

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