Simple Harmonic Motion

Holt Physics

Pages 438 - 451

Distinguish simple harmonic motion from other forms of periodic motion.

Periodic motion is motion in which a body moves repeatedly over the same path in equal time intervals.

Examples: uniform circular motion and simple harmonic motion.

Cont’d

Simple Harmonic Motion (SHM) is a special type of periodic motion in which an object moves back and forth, along a straight line or arc.

Examples: pendulum, swings, vibrating spring, piston in an engine.

In SHM, we ignore the effects of friction. Friction damps or slows down the motion of the

particles. If we included the affect of friction then it’s called damped harmonic motion.

Cont’d

For instance a person oscillating on a bungee cord would experience damped harmonic motion. Over time the amplitude of the oscillation changes due to the energy lost to friction.

http://departments.weber.edu/physics/amiri/director/DCRfiles/Energy/bungee4s.dcr

State the conditions necessary for simple harmonic motion.

A spring wants to stay at its equilibrium or resting position.

However, if a distorting force pulls down on the spring (when hanging an object from the spring, the distorting force is the weight of the object), the spring stretches to a point below the equilibrium position.

The spring then creates a restoring force, which tries to bring the spring back to the equilibrium position.

Cont’d

The distorting force and the restoring force are equal in magnitude and opposite in direction.

FNET and the acceleration are always directed toward the equilibrium position.

Cont’d

Applet showing the forces, displacement, and velocity of an object oscillating on a spring.

https://ngsir.netfirms.com/englishhtm/SpringSHM.htm

Displacement

Velocity

Acceleration

Cont’d

at equilibrium: – speed or velocity is at a maximum– displacement (x) is zero – acceleration is zero – FNET is zero (magnitude of Restoring

Force = magnitude of Distorting Force which is the weight)

– object continues to move due to inertia

Cont’d

at endpoints: – speed or velocity is zero – displacement (x) is at a maximum equal to the

amplitude – acceleration is at a maximum – restoring force is at a maximum

F= -kx (hook’s law)

– FNET is at a maximum

State Hooke’s law and apply it to the solution of problems.

Hooke’s Law relates the distorting force and the restoring force of a spring to the displacement from equilibrium.

Cont’d

F –restoring force in Newtons k – spring constant or force

constant (stiffness of a spring) in Newtons per meter (N/m)

x –displacement from equilibrium in meters

The distorting force is equal in magnitude but opposite in direction to the restoring force.

kxF

The spring shown to the right has an unstretched length of 3 cm. When a 2 kg object is hung from the spring, it comes to rest at the 7 cm mark. What is the spring constant of this spring?

490 N/m

What direction is the restoring force?

upward

Calculate the frequency and period of any simple harmonic motion.

T – period (time required for a complete vibration) in seconds

f – frequency in vibrations / second or Hertz

Tf

1

A particle is moving in simple harmonic motion with a frequency of 10 Hz.

What is its period? 0.1 sec How many complete oscillations does it

make in one minute? 600 oscilations

Relate uniform circular motion to simple harmonic motion.

The reference circle relates uniform circular motion to SHM.

The shadow of an object moving in uniform circular motion acts like SHM.

The speed of an object moving in uniform circular motion may be constant but the shadow won’t move at a constant speed.

The speed at the endpoints is zero and a maximum in the middle.

The shadow only shows one component of the motion.

Cont’d

Applet showing the forces, displacement, and velocity of an object oscillating on a spring and an object in uniform circular motion.

http://www.physics.uoguelph.ca/tutorials/shm/phase0.html

Identify the positions of and calculate the maximum velocity and maximum accelerations of a particle in simple harmonic motion.

The acceleration is a maximum at the endpoints and zero at the midpoint.

The acceleration is directly proportional to the displacement, x.

The radius of the reference circle is equal to the amplitude.

The force and acceleration are always directed toward the midpoint.

The mass on the end of a spring (which stretches linearly) is in equilibrium as shown. It is pulled down so that the pointer is opposite the 11 cm mark and then released.

What is the amplitude of the vibration?

4 cm

What two places will the restoring force be greatest?

11 cm and 3 cm

Where will the restoring force be least?

7 cm

Where is the speed greatest? 7 cm What two places is the speed

least? 3 cm and 11 cm

Where is the magnitude of the displacement greatest?

3 cm and 11 cm Where is the displacement least? 7 cm Where is the magnitude of the

acceleration greatest? 3 cm and 11 cm Where is the acceleration least? 7 cm Where is the elastic potential

energy the greatest? Where is the kinetic energy the

greatest?

Cont’d

Remember that in uniform circular motion, the velocity is calculated using:

vA

Tmax 2

T

rv

2

In SHM, the maximum velocity would be equal to the velocity of the object in uniform circular motion. The radius of the circle correlates to the Amplitude (A) in SHM.

Cont’d

Remember that in uniform circular motion, the centripetal acceleration is calculated using:

In SHM, the maximum acceleration would be equal to the acceleration of the object in uniform circular motion. The radius of the circle correlates to the Amplitude (A) in SHM.

A

va

2max

max r

vac

2

A mass hanging on a spring oscillates with an amplitude of 10 cm and a period of 2 seconds. What is the maximum speed of the object and where does it occur?

0.314 m/s at equilibrium What is the minimum speed of the object and where

does it occur? 0 m/s at the end points. What is its maximum acceleration? 0.987 m/s2 at the end points

An object moving is simple harmonic motion can be located using:

– A is amplitude– f is frequency– x is displacement from

equilibrium– ω is angular velocity

)cos(

)2cos(

tAx

or

ftAx

fT

22

The mass on the end of a spring (which stretches linearly) is in equilibrium as shown.

It is pulled down so that the pointer is opposite the 11 cm mark and then released. A spring vibrates in SHM according to the equation x = 4 cosπt.

How many complete vibrations does it make in 10 seconds?

5 vibrations

The elastic potential energy content of the system is 2

21 kxU s

So the maximum elastic potential energy is stored at the end points of the oscillations where the displacement is equal to the amplitude of the vibration

221 kAU s

At the end point, the object is not moving so there is no kinetic energy. Therefore the total energy content of the system is equal to

221 kAEo

A mass on a spring oscillates horizontally on a frictionless table with an amplitude of A. In terms of Eo (total mechanical energy of the system) when the mass is at A, Us = ______ and K = _________.

Us = Eo and K = 0When the mass is at 0.5 A, then Us = __________ and K

= _________. Us = 0.25 Eo and K = 0.75Eo

When the mass is at the equilibrium position, then Us = _________ and K = ________

Us = 0 and K = Eo

A 2 kg object is attached to a spring of force constant k = 500 N / m. The spring is then stretched 3 cm from the equilibrium position and released. What is the maximum kinetic energy of this system?

0.225 J What is the maximum velocity it will attain? 0.47 m/s

Cont’d

T = period (s)

m = mass (kg)

k = spring constant (N/m)T

m

k2

You want a mass that, when hung on the end of the spring, oscillates with a period of 3 seconds. If the spring constant is 5 N/m, the mass should be _______.

1.14 kg

The period for a mass vibrating on very stiff springs (large values of k) will be (larger / smaller) compared to the same mass vibrating on a less stiff spring.

Smaller If the value of k halves, the period will be ______

times as long.

2

Relate the motion of a simple pendulum to simple harmonic motion.

A pendulum is a type of SHM. A simple pendulum is a small, dense mass

suspended by a cord of negligible mass. The period of the pendulum is directly

proportional to the square root of the length and inversely proportional to the square root of the acceleration due to gravity.

Cont’d

T = period (s)

l = length (m)

g = acceleration due to gravity (m/s2)T

l

g2

A pendulum has a period of 2 seconds here on the surface of the earth. That pendulum is taken to the moon where the acceleration due to gravity is 1/6 as much. What is the period of the pendulum on the moon?

Squareroot of 6 times as much or 4.9 seconds.