Unformatted text preview: ω t ≈ ω t and cos ω t ≈ 1 − (ω t)2 /2. Therefore,
v0 cos θ
ω t = (v0 cos θ)t.
ω x(t) ≈ (96) And, using ω 2 = k/m,
y (t) ≈ −(k/m)t2
k + v0 sin θ
ω t = − gt2 + (v0 sin θ)t.
2 (97) These are the standard projectile results, as desired. For the above approximations to be valid, we need ω t
1 throughout the entire motion. If
we assume that ω (2v0 sin θ/g )
1, then the above approximations hold for
t = 2v0 sin θ/g , in which case we have y ≈ 0 at this time. That is, the projectile has hit the ground and the motion is ﬁnished. So “small ω ” means
g /(v0 sin θ).
Now consider large ω . For any t, the x(t) motion is simple harmonic. In order
CHAPTER 4. it to be in a straight
for the whole motion to be simple harmonic, we need OSCILLATIONS line, 32 so y/x must be a constant. This means that the (mg/k )(cos ω t − 1) term in
y (t) must be negligible. We therefore need
v0 sin θ
k =⇒ v0 sin θ
v0 sin θ =⇒ ω (98) This is what is meant by “large ω .” In this case, both x and y are (essentially)
proportional to sin ω t. The projectile reaches a maximum distance from the
origin of v0 /ω , and then heads back.
The above two conditions on ω can be summed up by saying that the time scale
of oscillations without gravity, namely 1/ω , should be much greater than or
much less than the time scale of projectile motion without the spring, namely
2v0 sin θ/g .
(c) We want y = 0 when x = 0. But x = (v0 cos θ) cos ω t, which is zero when
t = π /2. The y value at t = π /2 is (mg/k )(0 − 1) + (v0 sin θ/ω )(1). Setting
this equal to zero, and using k/m = ω 2 , gives g/ω 2 = v0 sin θ/ω =⇒ ω =
g/(v0 sin θ). This is, in a sense, right “between” the two limiting cases above.
4.23. Corrections to the pendulum
(a) F = ma in the tangential direction gives mg sin θ = mv dv/dx. Writing dx
as dθ , and separating variables and integrating gives
So v mg sin θ dθ =
θ0 dt = 0 =⇒ v = ± 2g (cos θ − cos θ0 ). (...
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