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# Let x be we found if someone claimed the rail

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Unformatted text preview: re the ball is moving perpendicular to the radial line from This checks in the special cases where C = 0 or D = 0. you to the ball, so therefore the velocity is never perpendicular to the slope of 4.16. Angledhill, which means that such a θ isn’t possible in the setup of Exercise 3.51, a rails as the position of each mass along that they could produce an angle position. Let x be we found. (If someone claimed the rail, relative to the equilibrium θ < θ0 in Then Exercise 3.51, then the-time reversed motion would yield an anglesin < θ0 for the spring stretches a distance 2x sin θ, yielding a force of 2kx θ θ. The which the distance decreases at is 2kx sin2 in the ﬂight, namely just after the component of this force along the rail some point θ. So F = ma along the rail gives ball passes ¨ through the “hill”, in contradiction with the result of this problem.) −2kx sin2 θ = mx. Hence, ω = 2k/m sin θ. 30 4.17. Eﬀective spring constant 3.53. Projectile with drag (a) Let the mass move a distance x to the right. Then the two springs pull to the (a) left= maforces −k= −αx −k2 x.y The−g − αy . Using the initial −(k1 + thexx F with gives x 1 x and˙ and ¨ = total force is therefore F = speed, k2 ) . ¨ ˙ equationeﬀ = k1 + k2 . Note that if k1 = 0, then keﬀ = k2 , as expected. And if integrates to Hence, k k1 = ∞, then keﬀ = ∞, as expected. x = Ae−αt =⇒ x = v0 cos θ e−αt . ˙ ˙ (53) (b) Let the mass move a distance x to the right. How much does each spring stretch? The initial position of springs must exert the same force, otherwise Assuming an key is that both zero, this then integrates to there would be a nonzero net force on some part of the massless springs, and this part= −(v0 cos θ/undergo+ B =⇒ x = (v0 cosLetαthe − e−αt ).stretch(54) by x would then α)e−αt inﬁnite acceleration. θ/ )(1 springs x1 and x2 . Then we have k1 x1CHAPTER 4. also x1 + x2 = x, of course. = k2 x2 . And OSCILLATIONS 29 Solving this system of two equations gives x1 = k2 x/(k1 + k2 ) and x2 = k1 x/(k1 + k2 ). The force in each spring (which is the force that the mass feels) is therefore k1 x1 = k2 x2 = k1 k2...
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## This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.

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