Physics 2101
Section
S cti n 3
March 31st
Announcements:
• Quiz today about Ch. 14
Class Website: http://www.phys.lsu.edu/classes/spring2010/phys2101‐‐3/
http://www.phys.lsu.edu/classes/spring2010/phys2101
http://www.phys.lsu.edu/~jzhang/teaching.html
“Simple Harmonic Motion (SHM)”‐‐‐Kinematics
x(t) = x max cos(ωt + ϕ )
ddx
v(t) =
= −x max ω sin (ωt + ϕ )
dt
dv
a(t ) =
= −x max ω 2 cos(ωt + ϕ )
dt
Position
Velocity
Acceleration
[v
max
= x max ω ]
2
a
=
x
ω
[ max max ]
a(t)
(t) = −ω 2 x(t)
(t)
In SHM, the acceleration is proportional to the displacement but opposite sign. The proportionality is the square of the angular it i
Th
ti
lit i th
f th
l
frequency. SHO and Hooke’s Law (again…)‐‐‐‐Dynamics
F = −kx
ma = −kx
k
a=− x
m
“In SHM, the acceleration is proportional to the displacement but opposite sign. The proportionality is h
l
the square of the angular frequency”. a(t) = −ω 2 x(t)
Block‐spring is linear SHO: k
k
→ ω=
m
m
ω
1
f =
→ f =
2π
2π
1
T = → T = 2π
f
ω2 =
NOTE: Independent of amplitude
k
m
m
k
Larger mass ‐> longer period
g
g p
SHO: relationships
x(t) = x max cos(ωt + ϕ )
shifted by −
π
2
v(t) = −x max ω sin (ωt + ϕ )
shifted by −
a(t) = −x maxω cos(ωt + ϕ )
π
v(t ) = 0 when x(t ) = ± xmax
v(t ) = ± vmax when x(t ) = 0
2
2
How to determine xmax and φ : INITIAL CONDITIONS
⎛ −v(0) ⎞
⎜
⎟ = tan (ϕ )
⎝ ω x(0) ⎠
v(0) = − xmaxω sin (ω (0) + ϕ )
Knowing x(0) and v(0), x(0) = xmax cos (ω (0) + ϕ )
x(0) = x max cos(ϕ )
x max =
x(0)
(0)
⎛ −1 ⎛ −v(0) ⎞⎞
cos⎜ tan ⎜
⎟⎟
⎝ ωx(0) ⎠⎠
⎝
Problem 15‐12: What is the phase constant for the harmonic oscillator with the velocity function v(t) shown in the figure, with x(t)
x(t) = xm cos(ω t + φ )
Answer:
v(t) = −xmω sin(ω t + φ )
vm = xmω
vs = v(0) = −vm sin(φ )
vs
4
sin(φ ) = − = − ;
vm
5
φ ≅ −53°
Example
A block whose mass m is 680 g is fastened to a spring whose constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium
pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0.
(a) What are the angular frequency, frequency, and period of the resulting motion?
f =
ω=
k
65 N/
N/m
=
= 9.8 rad/s
m
0.68 kg
T=
(b) What is the phase angle and amplitude of the oscillation? v0 = −x max ω sin (ω (0) + ϕ )
x 0 = x max cos(ω (0) + ϕ )
x(0) = x max cos(ϕ )
⎛
1
= 0.64 s
f
⎞
−0 m/s
⎟=0
⎝ (9.8 rad/s)( 0.11 m/s) ⎠
⎛ −v(0) ⎞
⎜
⎟ = tan (ϕ )
⎝ ω x(0) ⎠
x max =
ω
= 1.6 Hz
2π
ϕ = tan −1 ⎜
x(0)
= 11 cm
−1
cos(tan (0))
(c) What is the maximum speed and acceleration? vmax = xmaxω = (11 cm )( 9.8
9 8 rad/s
d/ ) = 1.1
1 1 m/s
/
amax = xmaxω2 = (11 cm)( 9.8 rad/s) = 11 m/s2
2
Energy swapping in SHO
The elastic property of the oscillating system (spring) stores potential energy
The inertia property (mass) stores kinetic energy: Total mechanical energy is constant
Knowing:
2
U(t) = 12 kx 2 = 12 kx max
cos 2 (ωt + ϕ )
2
KE(t) = 12 mv 2 = 12 mω 2 x max
sin 2 (ωt + ϕ )
k 2
x max sin 2 (ωt + ϕ )
m
2
= 12 kx max
sin 2 (ωt + ϕ )
= 12 m
E mech = U (t) + KE(t)
2
= 12 kxmax
(cos 2 (ωt + ϕ ) + sin 2 (ωt + ϕ ))
2
2
= 12 kxmax
= 12 mvmax
Exchange between KE
Exchange between KE and and U
U periodically during oscillation Clicker Question
Question 10‐3
A mass of is attached to one end of a spring and is free to slide on a A
mass of is attached to one end of a spring and is free to slide on a
frictionless horizontal surface. The mass is pulled back from its equilibrium position and released. What is the speed and kinetic energ of the mass hen it passes thro gh the eq ilibri m position
energy of the mass when it passes through the equilibrium position (x=0)?
1. Maximum
2 Zero
2.
3. Need more info
Problem 16‐17
m = 1 kg
A block of mass 1 kg (m) is on a horizontal surface (a shake table) g( )
(
)
that is moving back and forth horizontally with simple harmonic motion of frequency 2.0 Hz. (f) The coefficient of static friction between block and surface is 0.5 (μs).
μs = 0.5
f = 2 Hz
How great can the amplitude of the SHM be if the block is not to slip along the surface?
1) Angular frequency is: ω = 2 πf = 2 π (2 ⋅ Hz) = 12.6 ⋅ rad / s
2) Motion of shake table is: x(t) = x max cos(ωt )
→
a(t) = −x max ω 2 cos(ωt )
a max = x max ω 2
3) When the block just slips, it means the static friction is equal to the normal force: xˆ : − f s = ma SHM
yˆ : N − mg
g=0
f s,max = μ s N
= μ s mg
−μ s mg = − f s,max = ma SHM ,max
4) Putting together: ma SHM ,max = − μ s mg
a SHM ,max = − μ s g = x max ω 2
x max =
−μ s g
ω2
⎛ ((0.5)(9.8
)(
⋅ m /s2 ⎞
= ⎜⎜
= 0.031
0 031 ⋅ m
2 ⎟
⎟
⎝ (12.6 ⋅ rad / s ) ⎠
Problem15‐13: In the figure two springs of spring constant kl and kr are attached to a block of mass m. Find the frequency and period.
Let the positive x be to the right then the force on the block at x is:
kl
kr
− (xkl + xkr ) = ma
xkl + xkr )
(
2
a=−
= −ω a
m
f = 2πω
1
T=
f
Spring oscillations
t = 0
x(t) curves for the experiments involving a particular ()
p
g p
spring‐box systems oscillating in SHM (equal k’s). Rank (greatest‐to‐least) the curves according to:
a))
the system’s angular frequency
y
g
q
y
b)
the spring’s potential energy at t=0
c)
the box’s velocity at t=0
d)
the box’s kinetic energy at t=0
h b ’ ki i
0
e)
the box’s maximum kinetic energy
Problem 15‐35: A block of Mass M at rest is attached to a spring of constant k. A bullet of mass m and velocity v strikes and is embedded in the block. (a) find the speed of the block immediately after the collision: (b) The equation for the p
y
( )
q
displacement of the SHM.
( )
(a) Use conservation of momentum.
(M + m)v f = mv
vf =
x
mv
M +m
( )
(b) Find the equation of SHM.
q
x(t) = xm cos(ω t + φ )
At t = 0, x = 0 and v = vf
v(t) = −xmω sin(ω t + φ )
π
x(t ) = xm cos(ωt − )
2
π
v(t ) = −v f sin(ωt − )
2
And xm =
vf
ω