Supplementary HW #9: Ch. 15 1. The figure gives the position function of a 50g block in simple harmonic oscillation on the end of a spring. The period of the oscillation is 2 s. 1) Find the velocity and acceleration function? 2) Find the maximum kinetic energy, potential energy, and total energy. 3) Find the maximum force. 4) If you want to stop the oscillation in a quarter of period, what is minimum kinetic friction coefficient required? (Answer: μk = 0.02) Solutions: 2π
a) x(t ) = xm cos(ωt ) with ω =
= 3.14(1/ s) and xm = 0.07m
T
v(t ) = −ω xm sin(ωt )
a (t ) = −ω 2 xm cos(ωt )
kx 2
mv 2 m (ω xm )
kx 2
(b)
Fmax = −kxm ; U max = m ; K m = m =
= m
2
2
2
2
2
2π
4π m
m
(c ) T =
so that k =
= 2π
ω
k
T2
1
Since W = F ⋅ xm = μk mgxm = kxm2
2
2
1
1 4π m
2π 2
μk =
kxm =
( 2 ) xm = 2 xm = 0.02
2mg
2mg T
T g
2. Two springs with the same unstretched length but different force constants k1 and k2 are attached to a block of mass m on a level, frictionless surface. Calculate the effect force constant keff in each of the three cases (a), (b), and (c) depicted in the figure (the effective force constant is defined as ∑ Fx = − keff x .) 2
Solutions: (a) Image the mass moves a distance of Δx, the springs move Δx1 and Δx2
with force F1 = − k1Δx1 , F2 = − k2 Δx2
Δx = Δx1 = Δx2 , F = F1 + F2 = −(k1 + k2 )Δx
so keff = k1 + k2
(b) Since Δx = Δx1 − Δx2 ;
keff = k1 + k2
(c) F = F1 = F2 ; Δx = Δx1 + Δx2
so
Δx = −(
1 1
kk
+ ) F and keff = 1 2
k1 k1
k1 + k2
3. An approximation for the potential energy of a KCL molecule is 1
U = A[( R07 / 8r 8 ) − ] where R0 = 2.67 × 10‐10 m and A = 2.31 × 10‐28 J m. Using this r
approximation: (a) Show that radial component of the force on each atom 1
is Fr = A[( R07 / r 9 ) − 2 ] . Show that R0 is the equilibrium separation of the K and Cl atoms. r
(c) Find the minimum potential energy. (d) Use r = R0 + x and the first two terms of the binomial theorem { (1 + u ) n = 1 + nu + n(n − 1) u 2 + n(n − 1)(n − 2) u 3 + ⋅⋅⋅ , if u ≡ x/R0 <<1} to 2!
3!
show that Fr ≈ −(aA / R ) x so that the molecule’s force constant is k = 7 A / R03 . (e) With both the K and Cl atoms vibrating in opposite directions on opposite sides of the mm
molecule’s center of mass 1 2 = 3.06 × 10‐26 kg is the mass to use in calculating the m1 +m 2
frequency. Calculate the frequency of small‐amplitude vibration. (Answer: (e) f = 8.39 × 1012 Hz) Solutions: 3
0
(a) Fr = −
R7
1
dU
= A[( 90 ) − 2 ]
dr
r
r
(b) Set Fr = 0 for r = req
(c) U ( R0 ) = −
R07 1
that is, 9 = 2
req req
or req = R0
7A
= −7.57 ×10−19 J
8R0
(d) From (a) we have
R07
1
A r
r
Fr = A[( 9 ) − 2 ] = 2 [( ) −9 − ( ) −2 ]
r
r
R0 R0
R0
=
A
x
x
[(1 + ) −9 − (1 + ) −2 ] (x << R0 )
2
R0
R0
R0
≅
A
x
x
A
{1 − 9( ) − [1 − 2( )]} = −7 3 x
2
R0
R0
R0
R0
(e) f =
k/m
1
=
2π
2π
7A
= 8.39 ×1012 Hz
3
mR0
4. You want to construct a physical pendulum with a period of 4.0 s at a location where g = 9.8m /s2. (a) What is the length of a simple pendulum having this period? (b) Suppose the physical pendulum must mounted in a case that is not more than 0.5 m high. Can you devise a physical pendulum having a period of 4.0 s that will satisfy this requirement? (hint: consider a uniform slender rod of mass M and length L = 0.5 m and pivot the rod about an axis that is a distance d above the center of the rod) (Answer: (a) L =3.97 M; (b) d = 0.0053 m) Solutions: (a) T = 2π L / g so L = g (T / 2π ) 2 = 3.97m
(b) Use a uniform slender rod of mass M and length L = 0.5 m
Pivot the rod about an axis that is a distance d above the center of the rod.
The rod will oscillate as a physical pendulum with period T = 2π I / Mgd
Choose d so that T = 4.0 s
1
1
ML2 + Md 2 = M ( L2 + d 2 )
12
12
1
1 2
M ( L2 + d 2 )
L + d2
I
12
T = 2π
= 2π
= 2π 12
Mgd
Mgd
gd
I = I cm + Md 2 =
so that
T 2
) gd + L2 /12 = 0 (set L = 0.5 m and T = 4.0 s)
2π
d = 3.97 m or d = 0.0053 m
since d can not bigger than L/2, the correct answer is d = 0.0053 m = 0.53 cm
d2 −(
5. Two solid cylinders connected along their common axis by a short, light rod have radius R and total mass M and rest on a horizontal tabletop. A spring with force constant k has one end attached to a clamp and the other end to a frictionless ring at the center of mass of the cylinders (see the figure). The cylinders are pulled to the left a distance x, which stretches the spring, and released. There is sufficient friction between the tabletop and the cylinders to roll without slipping as they move back and forth on the end of the spring. (a) Show that the motion of the center of mass of the cylinders is simple harmonic. (b) Calculate its period in terms of M and k. (c) if the amplitude of the simple harmonic motion is A, what is the maximum value of the velocity of the center of mass of the cylinders. (Answer: (b) T = 2π 3M / 2k ) Solutions: (a) Apply ∑τ = I cmα and
∑F
x
= Macm
Based upon the coordination in the figure,
1
∑τ = fR = ( 2MR )α
∑ F = f − kx = − Ma
x
2
cm
1
1
MRα = Ma cm
2
2
so that
or f =
3
Macm (ax = −acm )
2
2k
ax = −
x thus give a SHO compared with ax = −ω 2 x
3M
kx =
(b) ω 2 = 2k / 3M
(c) U max =
vcm =
so that T = 2π / ω = 2π 3m / 2k
v
1 2
1
1
3
2
2
kA = KEmax = Mvcm
+ I cm ( cm ) 2 = Mvcm
2
2
2
R
4
2k
A
M
Another way of solving this problem form energy point of view:
Total mechanical energy is given as
2
ω2
Mv 2 I CM
Emech = U s +
+
2
2
2
dEmech d ⎛ 3Mvcm kx 2 ⎞ 3Mvcm ax
= ⎜
+
+ kxvcm
⎟=
dt
dt ⎝ 4
2 ⎠
2
However, as we know Emech = const. so that
⎛ 2k
ax = − ⎜
⎝ 3M
2k
⎞
⎟ x and ω =
3M
⎠
dEmech
=0
dt