Download H.P. Paar PHYS 4B: Mechanics, Fluids, Waves & Heat Spring 2015

H.P. Paar
PHYS 4B: Mechanics, Fluids, Waves & Heat
Spring 2015
Quizz # 1 solutions
Solutions by Yury Kiselev
1. (25 points)
(a) (10 points) Let’s set y = 0 at a position of the mass along vertical axis such that
the mass is at equilibrium. Let’s find how total force depends on displacement
y: we will use an axis looking up, so y > 0 means tha the mass is above the
equilibrium, so extra force produced by the upper spring is F1 = −yk1 . The
second spring is stretched when y > 0, so the force produced by the bottom spring
is also down: F2 = −yk2 . The gravitational force is constant and is cancelled by
the spring forces at the equilibrium position. So, F = ma, and
2
F = Fequilibrium + Fextra = 0 + (−yk1 ) + (−yk2 ) = ma = m ∂∂t2y ,
(1)
2
+k2
so ∂∂t2y + k1m
y = 0.
Of course, if we chose other y = 0 position, then we would have an additional
2
+k2
y + const. = 0., where const.
constant in our equation of motion: so ∂∂t2y + k1m
would depend on exact definition of y.
(b) (5 points) This is second order DE (the highest derivative is a second derivative),
linear (all factors involving the dependent variable (y) and its derivatives are
raisee to the first power (no logs or square-roots or whatever) and homogeneous
(no constant term), though we could write this equation relative to some other
point, instead of equilibrium point, and it would have a constant term, thus the
equation would be non-homogenous.
(c) (5 points) The most general solution can be written in different forms:
y(t) = A cos ωt + B sin ωt,
y(t) = A cos (ωt + φ0 ), or
y(t) = Aeiωt + Be−iωt ,
q
k1 +k2
m
as can be seen be substitution.
q
k1 +k2
(d) (5 points) Only ω =
is determined, but A and B or A and φ0 are not
m
determined, as we don’t know the initial conditions, which are usually an initial
coordinate and initial velocity.
where ω =
2. (25 points)
1
(a) (10 points) The system satisfies usual equation F = ma, where F = −kx and
2
2
k
x = 0. The solutions of this equation are cos ωt and sin ωt,
a = ∂∂t2x , so ∂∂t2x + m
q
q
k
ω
1
k
where ω = m
, so the frequency is f = 2π
= 2π
.
m
(b) (10 points) We can find amplitude of the oscillation by considering conservation
mv 2
of energy. After the push, energy of the system is E = U + K = 0 + 2 0 . At the
maximum stretch point (where x = A), energy is E = U + K = kA2 /2 + 0. They
are equal, because there is no damping (surface
q is frictionless and we neglect air
resistance), and so kA2 /2 = mv02 /2, or A = v0
m
.
k
(c) (5 points) No, it is not damped, the surface is frictionless and we usually neglect
air resistance.
3. (50 points)
(a) (10 points) Total torque is τtotal = −Kθ + τ0 cos ωt = Iα, where I is the moment
2
2
of inertia and α = ∂∂t2θ the angular acceleration. This yields I ∂∂t2θ + Kθ = τ0 cos ωt.
(b) (5 points) This is second order DE, linear, but not homogeneous.
(c) (20 points) We can find the solution by substituting θ = θ0 + θw , where θ0 is
2
a general solution of a homogeneous equation I ∂∂tθ20 + Kθ0 = 0 and θw is any
2
solution of I ∂∂tθ2w + Kθw = τ0 cos ωt. Then their sum θ will satisfy the original
equation. Let’s guess form of θw : θw = A cos ωt and substitute it into the equation:
2
I ∂∂tθ2w + Kθw = IAω 2 (− cos ωt) + KA cos ωt = τ0 cos ωt, so this means by choosing
τ0
A we can find one solution for θw : A = K−Iω
2 . We can combine it with the general
q
solution of homogeneous equation B cos ω0 t + C sin ω0 t, where ω0 = KI to get
the general solution:
θ=
τ0
I(ω02 −ω 2 )
cos ωt + B cos ω0 t + C sin ω0 t,
(d) (5 points) B and C are impossible to specify, because initial conditions are not
given. Other constants are already specified.
(e) (10 points) Resonance behavior means that the amplitude of the oscillations becomes large at a specific frequency (the resonance frequency) of the applied force.
We can see that the first term indeed shows resonance at ω = ω0 . When ω = ω0
the amplitude goes to infinity. This is not physical: it happens only because
we neglected damping. Even if we could neglect damping at original velocities,
when amplitude grows, characteristic velocities grow too and damping can not be
neglected.
4. (25 points) If the train moves with constant velocity, it’s still an inertial system, so the
physical laws are the same as is in
qa train at rest (Newton’s first Law). The frequency
didn’t change and remains ω =
g
.
l
The reason why it oscillates is not important –
2
it’s either have residual momentum from initial push, or it may oscillate because of
small bumps of the rails.
3