Transcript Chapter 10 Lesson 2

Simple Harmonic Motion
Lesson 2
Work Done in Stretching a Spring
Work done ON the spring is positive;
work BY spring is negative.
x
From Hooke’s law the force F is:
F (x) = kx
F
F
m
To stretch spring from
x1 to x2 , work is:
Work  ½kx  ½kx
2
2
x1
x2
2
1
Example 2: A 4-kg mass suspended from a
spring produces a displacement of 20 cm.
What is the spring constant?
The stretching force is the weight
(W = mg) of the 4-kg mass: 20 cm
F = (4 kg)(9.8 m/s2) = 39.2 N
F
m
Now, from Hooke’s law, the force
constant k of the spring is:
k=
DF
Dx
=
39.2 N
0.2 m
k = 196 N/m
Example 2(cont.: The mass m is now stretched
a distance of 8 cm and held. What is the
potential energy? (k = 196 N/m)
The potential energy is equal to
the work done in stretching the
spring:
Work  ½kx  ½kx
2
2
8 cm
0
m
2
1
U  ½kx  ½(196 N/m)(0.08 m)
2
U = 0.627 J
F
2
Displacement in SHM
x
m
x = -A
x=0
x = +A
• Displacement is positive when the position is
to the right of the equilibrium position (x = 0)
and negative when located to the left.
• The maximum displacement is called the
amplitude A.
Velocity in SHM
v (-)
v (+)
m
x = -A
x=0
x = +A
• Velocity is positive when moving to the right
and negative when moving to the left.
• It is zero at the end points and a maximum
at the midpoint in either direction (+ or -).
Acceleration in SHM
+a
-x
+x
-a
m
x = -A
x=0
x = +A
• Acceleration is in the direction of the
restoring force. (a is positive when x is
negative, and negative when x is positive.)
F  ma  kx
• Acceleration is a maximum at the end points
and it is zero at the center of oscillation.
Acceleration vs. Displacement
a
v
x
m
x = -A
x=0
x = +A
Given the spring constant, the displacement, and
the mass, the acceleration can be found from:
F  ma  kx or
 kx
a
m
Note: Acceleration is always opposite to displacement.
Example 3: A 2-kg mass hangs at the end
of a spring whose constant is k = 400 N/m.
The mass is displaced a distance of 12 cm
and released. What is the acceleration at
the instant the displacement is x = +7 cm?
 kx
a
m
(400 N/m)(+0.07 m)
a
2 kg
a = -14.0 m/s2
a
m
Note: When the displacement is +7 cm
(downward), the acceleration is -14.0 m/s2
(upward) independent of motion direction.
+x
Example 4: What is the maximum acceleration
for the 2-kg mass in the previous problem? (A
= 12 cm, k = 400 N/m)
The maximum acceleration occurs
when the restoring force is a
maximum; i.e., when the stretch or
compression of the spring is largest.
F = ma = -kx
xmax =  A
kA 400 N(  0.12 m)
a

m
2 kg
Maximum
Acceleration:
m
amax = ± 24.0 m/s2
+x
Conservation of Energy
The total mechanical energy (U + K) of a
vibrating system is constant; i.e., it is the
same at any point in the oscillating path.
a
v
x
m
x = -A
x=0
x = +A
For any two points A and B, we may write:
½mvA2 + ½kxA 2 = ½mvB2 + ½kxB 2
Energy of a Vibrating System:
A
x
a
v
m
x = -A
x=0
B
x = +A
• At points A and B, the velocity is zero and the
acceleration is a maximum. The total energy is:
U + K = ½kA2 x =  A and v = 0.
• At any other point: U + K = ½mv2 + ½kx2
Velocity as Function of Position.
x
a
v
m
x = -A
1
2
x=0
k
v
A2  x 2
m
mv  kx  kA
2
1
2
2
1
2
vmax when
x = 0:
x = +A
2
v
k
A
m
Example 5: A 2-kg mass hangs at the end of
a spring whose constant is k = 800 N/m. The
mass is displaced a distance of 10 cm and
released. What is the velocity at the instant
the displacement is x = +6 cm?
½mv2 + ½kx 2 = ½kA2
k
v
A2  x 2
m
800 N/m
v
(0.1 m) 2  (0.06 m) 2
2 kg
v = ±1.60 m/s
m
+x
Example 5 (Cont.): What is the maximum
velocity for the previous problem? (A = 10
cm, k = 800 N/m, m = 2 kg.)
The velocity is maximum when x = 0:
0
½mv2 + ½kx 2 = ½kA2
v
k
800 N/m
A
(0.1 m)
m
2 kg
v = ± 2.00 m/s
m
+x
The Simple Pendulum
The period of a simple
pendulum is given by:
L
T  2
g
L
For small angles q.
1
f 
2
g
L
mg
Example 8. What must be the length of a
simple pendulum for a clock which has a period
of two seconds (tick-tock)?
L
T  2
g
2
2 L
T  4
;
g
(2 s) 2 (9.8 m/s 2 )
L
2
4
L
T 2g
L=
2
4
L = 0.993 m
To Be Continued…