Lecture 21: Ideal Spring and Simple Harmonic Motion

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Lecture 21: Ideal Spring and Simple Harmonic Motion. New Material: Textbook Chapters 10.1, 10.2 and 10.3 . relaxed position. F X = 0. x. x=0. Ideal Springs.

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Lecture 21: Ideal Spring and Simple Harmonic Motion
• New Material: Textbook Chapters 10.1, 10.2 and 10.3

relaxed position

FX = 0

x

x=0

Ideal Springs
• Hooke’s Law:The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position (for small x).
• FX = -k xWhere xis the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)
Ideal Springs
• Hooke’s Law:The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
• FX = -k xWhere xis the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)

relaxed position

FX = -kx > 0

x

• x  0

x=0

Ideal Springs
• Hooke’s Law:The force exerted by a spring is proportional to the distance the spring is stretched or compressed from its relaxed position.
• FX = -k xWhere xis the displacement from the relaxed position and k is the constant of proportionality. (often called “spring constant”)

relaxed position

FX = - kx < 0

x

• x > 0

x=0

Simple Harmonic Motion

Consider the friction-free motion of an object attached

to an ideal spring, i.e. a spring that behaves according to

Hooke’s law.

How does displacement, velocity and acceleration

of the object vary with time ?

Analogy:

Simple harmonic motion along x

<-> x component of uniform circular motion

x = R cos q

=R cos (wt)

sinceq = wt

x

1

1

R

2

2

3

3

0

y

4

6

-R

4

6

5

5

What does moving along a circular path have to do with moving back & forth in a straight line (oscillation about equilibrium) ??

x

8

8

q

R

7

7

Velocity and Acceleration
• Using again the reference circle one finds for the velocity

v = - vT sin q = - A w sin (w t)

and for the acceleration

a = - ac cos q = - A w2 cos (w t)

x

+A

CORRECT

t

-A

Concept Question

A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest?

1. When x = +A or -A (i.e. maximum displacement)

2. When x = 0 (i.e. zero displacement)

3. The speed of the mass is constant

x

+A

CORRECT

t

-A

Concept Question

A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the magnitude of the acceleration of the block biggest?

1. When x = +A or -A (i.e. maximum displacement)

2. When x = 0 (i.e. zero displacement)

3. The acceleration of the mass is constant

Springs and Simple Harmonic Motion

X=0

X=A; v=0; a=-amax

X=0; v=-vmax; a=0

X=-A; v=0; a=amax

X=0; v=vmax; a=0

X=A; v=0; a=-amax

X=-A

X=A

Simple Harmonic Motion:

At t=0 s, x=A or At t=0 s, x=0 m

x(t) = [A]cos(t)

v(t) = -[A]sin(t)

a(t) = -[A2]cos(t)

x(t) = [A]sin(t)

v(t) = [A]cos(t)

a(t) = -[A2]sin(t)

OR

Period = T (seconds per cycle)

Frequency = f = 1/T (cycles per second)

Angular frequency =  = 2f = 2/T

For spring: 2 = k/m

xmax = A

vmax = A

amax = A2

Elastic Potential Energy
• Work done by the (average) restoring force of the spring is

W = |Fave| s cos q = ½ k ( x0+xf) (x0-xf) =

= ½ k (x02 – xf2) = Epot,elastic,0- Epot,elastic,f

The elastic potential energy has to be considered when

Calculating the total mechanical energy of an object

attached to a spring.