In this blog post, we will discuss how to solve the problem of “What Will Be the Displacement δy at Point x=2.5m?” The displacement δy is given by: δy = ∫0t (uy)dt To solve for y(x, t), use integration by parts as follows: ∫0t (v+w)dt = v(1-u2)+w'(1-u2)−∫0t u du and then take the derivative with respect to t from both sides. This gives us that w’=-du/dv and so our equation becomes that y(x, t)=uv(x-ut)+(∫0t u du)/(v+w). This equation can be integrated by substitution to give: y = ∫0t v′du/dv−u ln|∫0t w′du/dv|. Finally, we take the natural logarithm of both sides and find that y=ln|∫0t w′du/dv|. This means that at time t=T (or x=L), our displacement δy is equal to -ln(L)δyt where L represents the length of the spring in meters. If you want a calculator showing this result for values other than T or L just go here: Google Calculator: What Will Be the Displacement δy at Point x=L? Examples of Application in Engineering: A spring is often used for energy storage. As we know, when a force F is applied to an elastic object it stores potential energy equal to -kx where k represents how “stiff” or “soft” the material that make up the object. The displacement δy can be used as a metric by which engineers measure stiffness. If y(x)=-ln|∫0t w′du/dv| then this means that as v approaches infinity and so does constant speed (in other words, velocity), then y will approach zero also because L becomes infinite;

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