The mass shown in figure 1 is resting on a frictionless horizontal table (meaning that it...

  

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The mass shown in figure 1 is resting on a frictionless horizontal table (meaning that it can move in the plane of the sheet, not just along the axis shown on the figure). Each of the two identical springs has stiffness k and unstretched length lo. At equilibrium the mass rests at the origin, and the distances are not necessarily equal to lo. (That is, the spring might already be stretched or compressed.) 1.1. Show that when the mass moves to a position (x, y), with x and y small, the potential energy with respect to the x = 0, y = 0 equilibrirum has the form AU= = 1/2 (kxx + kyy), (1) i.e., that of an anisotropic oscillator (anisotropic = behaves differently depending on the di- rection). Hints: this one is difficult. "Small" does indeed mean that you need to do a Taylor expansion, but you will not be able to stop after the first order, you will need to go further. You may need the following results: For < < < 1, 1+l+ 2 8 (a+b+c) = a + b + c + 2ab +2ac+2bc. (3) Finally, every time your Taylor expansions turn up a term of order that is higher than x, y or xy, discard it. 1.2. Show that if l k Jeeeven m veeeeel k The mass shown in figure 1 is resting on a frictionless horizontal table (meaning that it can move in the plane of the sheet, not just along the axis shown on the figure). Each of the two identical springs has stiffness k and unstretched length lo. At equilibrium the mass rests at the origin, and the distances are not necessarily equal to lo. (That is, the spring might already be stretched or compressed.) 1.1. Show that when the mass moves to a position (x, y), with x and y small, the potential energy with respect to the x = 0, y = 0 equilibrirum has the form AU= = 1/2 (kxx + kyy), (1) i.e., that of an anisotropic oscillator (anisotropic = behaves differently depending on the di- rection). Hints: this one is difficult. "Small" does indeed mean that you need to do a Taylor expansion, but you will not be able to stop after the first order, you will need to go further. You may need the following results: For < < < 1, 1+l+ 2 8 (a+b+c) = a + b + c + 2ab +2ac+2bc. (3) Finally, every time your Taylor expansions turn up a term of order that is higher than x, y or xy, discard it. 1.2. Show that if l k Jeeeven m veeeeel k

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Solutions given data The mass shown from fig ... View the full answer