# Tag Info

1

Your approach ignores the body of the slinky and essentially describes two massive particles coupled by a very light spring, which is not allowed to oscillate or show any of the interesting dynamics a real slinky will exhibit. Ideally, you should be using some sort of continuum-mechanics approach to this problem, e.g. treating the slinky as a very elastic ...

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Just try this: Tie a spring to a wall and pull it. Do you feel a resistive force? Or do you feel a force that is somewhat helping the force that you are applying? In other words, Does the spring pull on you or push you away when you pull on it? It pulls you, of course, and this is a direct result from Newton's third law, which in this case gives: ...

1

If the force exerted by the spring on the attached object / the acceleration of the object is in the same direction as its displacement, you can imagine that the object will continue to go to infinity because there is no opposite force bringing the object back to the equilibrium position. Hence, the minus sign give us the sense that the acceleration of the ...

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$x$ measures the difference in length of the spring in relation to its relaxed state. If you increase the length (positive $x$), the spring creates a force in the negative $x$ direction, because it wants to return to its relaxed state. Accordingly, if you compress the spring (negative $x$) the spring wants to expand (force in positive $x$ direction) in order ...

2

Two springs in parallel effectively behave as a single spring with spring constant $k_\mathrm{parallel} = k_1+k_2 = 2k$ while two springs in series effectively behave as a single spring with spring constant $k_\mathrm{series} = k_1k_2/(k_1+k_2) = k/2$. Recall that the period of a mass on a spring is $$T = 2\pi \sqrt{\frac{m}{k}}$$ Which gives $$... 0 You are using the wrong law . This is linear SHM. Find out the time period for that , it will be 2\pi \sqrt \frac{m}{k} . Then figure 1 , springs are in parallel and in figure 2 , springs are in series . Do a wiki search to figure out how to find equivalent springs . An Advice : Don't learn equations blindly . 0 Max. Energy in spring=\dfrac12 kx^2 that will get transfered into body of mass m=65kg . So,$$\dfrac12 kx^2=\dfrac12 mv^2$$here you get the velocity and then solve for range$$R\text{ange}=\dfrac{u^2sin(2\theta)}{g}=\dfrac{u^2}{g}\Bigg|_{\theta=45}$$1 Actually, this is just a simple model for resistive forces (usually duo to viscosity) , and in many situations you can not assume this force to be linearly dependent on velocity.(for example ,usually this model is correct only for small enough objects) In many everyday examples, this force is due to viscous forces. If you consider an (small enough) object ... 3 This question is actually one of the lab exercises I teach. For a spring-mass system, if the damping force is friction, then it is independent of velocity (verified experimentally). However, as mentioned in the comments, the damping force may not always be friction. For example, if the mass is a material like aluminium and it is oscillating over some ... 0 Use Reduced Mass to tackle such problems. Now just see this as a simple spring mass system with mass as reduced mass like this one. Where the hanging mass is equivalent reduced mass. 0 When the acceleration is a function of position use the following$$ a(x) = \frac{{\rm d}v}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} \frac{{\rm d}x}{{\rm d}t} = \frac{{\rm d}v}{{\rm d}x} u  \int a(x)\,{\rm d} x = \int u\,{\rm d} u = \frac{1}{2} u^2 + K_1 $$which is solved for u(x). The the position is found from$$ t = \int \frac{1}{u(x)}\,{\rm d} ...

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You certainly want to numerically integrate your motion equation givent your expression of the force. I will assume that the mass $m$ is attached at one side of the rope, while the other side is attached to a wall or something that won't move during the integration. Something like this, where the spring is in fact your rope, with $x$ the extension of the ...

2

Given a spring with spring constant $k$ whose extension is in the direction $x$, the magnitude of the force that the spring exerts is given by $$|F| = k|L-l|$$ where $L$ is its length and $l$ is its equilibrium length. Now, imagine that the two masses are at positions $x_1$ and $x_2$ with $x_2>x_1$, then the length of the spring is given by \$L = x_2 ...

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