Simple Harmonic Motion

Spring motion

Let’s assume that a mass is attached to a spring, pulled back, and allowed to move on a frictionless surface…

Simple Harmonic Motion (SHM)

We know that if we stretch a spring with a mass on the end and let it go, the mass will oscillate back and forth (if there is no friction).

This oscillation is called Simple Harmonic Motion, and is actually very easy to understand...

km

km

km

SHM Dynamics

At any given instant we know that F = ma must be true.But in this case F = -kx and F = ma

So: -kx = ma F = - kx (Hooke’s

Law) Period is proportional to

the square root of mass over spring constant

2m

Tk

k

x

m

F = -kx a

km

T 2

2 F kx k m

SHM Dynamics...

y = R cos = R cos (t)

But wait a minute...what does angular velocity have to do with moving back & forth in a straight line ??

x

y

-1

1

0

1 12 2

3 3

4 45 5

6 62

SHM and Velocity and Acceleration

If you were to plot the position of the mass over time, it would look like a sine wave.

The velocity and acceleration would look a little out of phase

Problem: Vertical Spring

A mass m = 102 g is hung from a vertical spring. The equilibrium position is at y = 0. The mass is then pulled down a distance d = 10 cm from equilibrium and released at t = 0. The measured period of oscillation is T = 0.8 s.– What is the spring constant k?

k

m

y

0

-d

t = 0

The Simple Pendulum...

Period is proportional to the square root of the length over gravity

This works when the angle is less than 15o

The period does not depend on the mass of the object

L

dm

mg

z

TLg

2

Simple Harmonic Motion

You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1.

Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2.

Which of the following is true:

(a) T1 = T2

(b) T1 > T2

(c) T1 < T2

Solution

L1

L2

Standing up raises the CM of the swing, making it shorter!

T1 T2

Since L1 > L2 we see that T1 > T2 .

Energy in SHM

For both the spring and the pendulum, we can derive the SHM solution by using energy conservation.

The total energy (K + U) of a system undergoing SHM will always be constant!

This is not surprising since there are only conservative forces present, hence K+U energy is conserved.

-A A0 s

U

U

KE

Simple Harmonic Motion

The End