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## Hot answers tagged spring

33

If the spring locks when the ball is at rest in the lab frame, then by the arguments you give it follows that the spring must not be compressed at all. This is indeed the case. As the ball slows down, the block begins to speed up. Eventually they are traveling at the same speed, at which point the spring has reached its maximum compression. As the spring ...

7

The spring constant depends on the material and geometry and not a net effect of other environmental factors (ignoring temperature dependence). It measures the stiffness of the spring.

6

I was just wondering how the value of a spring constant would change if gravity were increased. Gravity has nothing to do with the spring constant. It can only effect the net force on the spring depending on the orientation of the spring. Agreeing that gravity is 9.8 m/𝑠2, if, for example, gravity was changed to 15 m/𝑠2 with nothing else to consider in ...

4

As @mikestone suggested in the comments, the easiest way to solve this problem is by summing the potential energies of all the springs. However, I would like to comment on the approach described in the question: the equations of motion can be written as $$m\ddot{x}_1 = -\frac{\partial V(x_1, x_2)}{\partial x_1} = -2kx_1 + kx_2,\\ m\ddot{x}_2 = -\frac{\... 3 If the spring locks when the deformation is maximum the collision is not ellastic. At that point the relative velocity of the two objects is zero so they continue to move with the same velocity. This is a model of non-elastic collision. The collisions are ellastic when the spring does not lock and restore its potential energy into kinetic. For actual balls ... 3 The problem is exactly how billiard balls work (in 1 dimension, it's a much easier game like that). The spring is replaced by the elasticity of the ball, which doesn't need to lock, because that's how billiards works: all the cue ball's E and p go to the target ball, leaving the cue ball at rest. 3 Question Can be Solved without Integration. Potential Energy is stored in the springs only. Let The Rightmost spring is stretched by amount  x_2, And the Leftmost spring is stretched by amount  x_1 then the Middle spring will be stretched by amount (x_2 - x_1).(As\ shown\ in \ figure )$$(2a +x_2)-(a+ x_1)= a+ x_2 - x_1 \qquad (Proof)$$where a ... 2 After the collision, the two blocks have the same velocity v'. Conservation of momentum: mv = 2mv', so v' = v/2. Conservation of energy: \frac{1}{2} m v^2 = \frac{1}{2}(2m)v'^2 + PE_{spring} = \frac{1}{4}mv^2 + PE_{spring} So, PE_{spring} = \frac{1}{4}mv^2. (Half the incident energy gets stored in the spring.) 1 You seem to be confused with the superposition of two oscillators. The following discussion will be useful for you : Waves are usually described by variations in some parameters through space and time—for example, height in a water wave, pressure in a sound wave, or the electromagnetic field in a light wave. The value of this parameter is called the ... 1 Yes, typical bathroom scales measure weight mg, not mass m. But for convenience they are calibrated to show a reading in mass units, eg kilograms (or pounds). So if the scales indicate that your "weight" is 50 kg that really means that your mass is 50 kg. Your weight is your mass times the gravitational acceleration, which is approximately 10 m/... 1 Too long for a comment. It's not clear what is troubling you. The rightmost spring does not communicate with M_1; it affects the displacement x_2 of M_2, which is already factored in the extension or contraction of the middle spring. If you assume x_1=x_2, the middle spring has not changed its equilibrium length, so it does not exchange any force ... 1 I) Equations of motion Kinetic energy :$$T=\frac{m}{2}\left(\dot{x}^2_1+\dot{x}_2^2\right)$$Potential energy$$U=\frac{k}{2}\left(x_1^2+(x_2-x_1)^2+x_2^2\right)$$with Euler Langrage you get:$${\ddot x}_{{1}}+{\frac {2\,kx_{{1}}-kx_{{2}}}{m}}=0\tag 1{\ddot x}_{{2}}+{\frac {2\,kx_{{2}}-kx_{{1}}}{m}}=0\tag 2II) Equations of motion: Normal mode In ... 1 You should notice that the other normal coordinate is implied to be fixed at zero while you consider the motion along the normal coordinate n_1. The normal coordinates of two particles (or blocks in this case) can generally be written as \begin{align} n_1 =& a_{11} x_1 + a_{12} x_2, \\ n_2 =& a_{21} x_1 + a_{22} x_2.\label{eq: n1n2}\tag{1} \end{... 1 First, as G. Smith, mentioned, the kinetic energy term should have \dot{\theta}^2 instead of \ddot{\theta}^2. I'm assuming that's a typo. Your mistake here is in your choice of pivot. The problem statement claims that "the string to the right is inextensible." That means that the pivot should be chosen as the point on the right where the string ... 1 Here's a slightly different take on someone else's answer that does not assume the spring constants are necessarily the same. It also uses the inspection method, based on the idea that the potential energy of a spring is \frac{1}{2}\kappa \times \hbox{stretch}^2 (or compression). For your first spring on the left, the stretch would come from the ... 1 The difference is in the coordinate system. You are doing your calculation with x=0 as the point where the string is in its natural length. The equilibrium position, in that case, is (m+m_c/2) g/k. However, your professor is taking that equilibrium position x_o=(m+m_c/2) g/k as his origin x' = 0 in the shifted coordinate system. You can arrive at ... 1 In this example, we have to assume that the spring is is attached to the block in such a way that it allows free rotation of lower end of the spring as if it is hinged to the block. In this case the spring will apply force longitudinally so that the elongation produced in the spring will be equal to the difference between its final and initial lengths. This ... 1 In the formula W= \frac {kx^2}{2}, the x is the extension in the spring, not the displacement of the block. As the spring is pivoted and free to rotate, even if the displacement of the block is making some angle with the force, the extension in the spring is always parallel to the force. Thus, the assumptions made while deriving the formula still hold ... 1 So the potential energy of a spring oriented along the z-axis is U=\frac12 k(z-z_0)^2  where $z_0$ is the position that the spring occupies at rest. Note that the spring constant $k$ has to deal with how sharp the parabola is, but the force depends on not just the spring constant but also the displacement. When we add gravity along the $z$-axis we ...

1

Assuming you're using the same mass to measure $k$ (spring constant) under different gravitational accelerations, $k$ will always remains the same. With a greater acceleration due to gravity than where you originally measured the spring constant, the forces acting upon the spring will be different. However, that doesn't mean that the spring constant is ...

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