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# So the general solution for the displacement z from

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Unformatted text preview: x/(k1 + k2 ), directed to the left. Therefore, keﬀ = k1 k2 /(k1 + k2 ). Note that if k1 = 0, then keﬀ = 0, as expected. And if k1 = ∞, then keﬀ = k2 , as expected. 4.18. Changing k Let’s ﬁrst ﬁnd the new equilibrium position. If it is a distance d to the right of the center, then we have k( + d) = 3k( − d) =⇒ d = /2. The eﬀective spring constant is k + 3k = 4k, because moving the mass a distance y to the right changes the force from the left spring by −ky , and also changes the force from the right spring by −3ky . So the general solution for the displacement, z , from equilibrium is z (t) = A cos 2 k/m t + B sin 2 k/m t . (90) The initial conditions z (0) = − /2 and v (0) = 0 quickly give A = − /2 and B = 0. So we have z (t) = −( /2) cos(2 k/m t). Adding on the /2 for the equilibrium position gives the position relative to the center as x(t) = ( /2) 1 − cos(2 k/m t) . if it were massless. Now cut the string connecting the springs. The two limp strings acquire a tension, and we now have two springs in parallel, instead of in series; see the second setup in Fig. 4. Each spring needs to support only half the weight, so each one is stretched by mg/2k. The mass therefore hangs a distance mg/2k below where it would hang if it were massless. This is 3mg/2k above where it hung before the string was cut. So the mass does indeed rise. Strange but true. 4.22. Projectile on a spring ˆ (a) The force on the projectile is F = −kr − mg y. The x component of F = ma is therefore mx = −kx =⇒ x(t) = A cos ω t + B sin ω t, with ω = k/m. And ¨ the y component is my = −ky − mg =⇒ mz = −kz , where z ≡ y + mg/k . So ¨ ¨ z takes the standard trig form, which yields y (t) = C cos ω t + D sin ω t − mg/k . The initial conditions x(0) = 0 and x(0) = v0 cos θ quickly give ˙ x(t) = v0 cos θ sin ω t. ω (94) And the initial conditions y (0) = 0 and y (0) = v0 sin θ give ˙ y (t) = mg k (cos ω t − 1) + v0 sin θ sin ω t. ω (95) (b) For small ω t, we have sin...
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