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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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 Fall '13
 Brandenberger
 mechanics, Force

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