In the expression
A==Fm
m√
(ω^2 −ωo^2 )^2 +ω^2 oω^2 Q−^2from page 1024, substitutingω=ωomakes the first term inside the square root vanish, which
should make the denominator pretty small, thereby producing a pretty big amplitude. In the
limit ofQ=∞,Q−^2 = 0, so the second term vanishes, andω=ωoactually produces an infinite
amplitude. For values ofQthat are large but finite, we still expect to get resonance pretty close
toω=ωo. Settingω=ωoin the finite-Qcase, the first term vanishes, we can simplify the
square root, and the result ends up beingA∝ 1 /√
Q−^2 ∝Q. This is only an approximation,
because we had to assume early on thatQwas large.
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F=ma
Page 199:Answers to self-checks for chapter 4
Page 262:
Torques 1, 2, and 4 all have the same sign, because they are trying to twist the wrench clockwise.
The sign of 3 is opposite to the signs of 1, 2, and 4. The magnitude of 3 is the greatest, since it
has a largerand the force is nearly all perpendicular to the wrench. Torques 1 and 2 are the
same because they have the same values ofrandF⊥. Torque 4 is the smallest, due to its small
r.
Page 271:
One person’sθ-tgraph would simply be shifted up or down relative to the others. The derivative
equals the slope of the tangent line, and this slope isn’t changed when you shift the graph, so
both people would agree on the angular velocity.
Page 273:
Reversing the direction ofωalso reverses the direction of motion, and this is reflected by the
relationship between the plus and minus signs. In the equation for the radial acceleration,ωis
squared, so even ifωis negative, the result is positive. This makes sense because the acceleration
is always inward in circular motion, never outward.
Page 285:
All the rotations around thexaxis giveωvectors along the positivexaxis (thumb pointing along
positivex), and all the rotations about theyaxis haveωvectors with positiveycomponents.
Page 288:
For example, if we take (A×B)x=AyBz−ByAzand reverse the A’s and B’s, we get (B×A)x=
ByAz−AyBz, which is just the negative of the original expression.
Answers to self-checks for chapter 5
Page 311:
Solids can exert shear forces. A solid could be in an equilibrium in which the shear forces were
canceling the forces due to unequal pressures on the sides of the cube.