SIMPLE HARMONIC
MOTION PROBLEMS (RD SEC 12-1, 12-2 first)

Simple Harmonic
Oscillators/Waves/

Pendulum Period=

Spring: Period=

where k is the
spring constant k= Force/distance
= ma/x

Period T =1/ƒ , ƒ
= 1/T, v = ƒ * WL for any wave

***x = A_{0} sin ω *t* where ω^{2} = k/m , ω=
angular frequency = 2*π *ƒ

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1 A clown is
rocking on a rocking chair in the dark. His glowing red nose moves back and
forth a distance of 0.42 m exactly 30 times a MINUTE, in a simple harmonic motion.

Draw a picture

a) What is the
amplitude of this motion? * Amplitude is .21 m*

b) What is the
period of this motion? * 30
times a minute…*

*Period (T) =
time/wave *

*= 1 min/30
times = 60 sec/30 times = 2 sec/wave*

c) What is the
frequency of this motion?

*f = wave/time = 1 wave/2sec=.5 Hz*

d. The top of the
clown’s hat contains a small light bulb that shines a narrow light beam. The
beam makes a spot on the wall that goes back and forth between two dots placed
1.00 m apart as the clown rocks. What are the amplitude, period, and frequency
of the motion?

*Period and
Frequency are the same! (T = 2 sec, f= .5 Hz) *

*Amplitude is
0.5 m*

2) a 5.00 kg block
hung on a spring causes a 10.0 cm elongation of the spring.

a) What is the
restoring force exerted on the block by the spring?

*F restoring
= Weight = mg = 5 (9.8) = 49 N*

b) What is the
spring constant? *K = F/x = 49 / (.1 m) = 490 N/m*

c) What force is
required to stretch this spring 8.5 cm horizontally?

*F=kx
= 490 (.085)=41.65 N*

d) What will the
spring’s elongation be when pulled by a force of 77.7 N?

*F=
k x*

* 77.7
= 490 x*

* x
= .1586 m = 15.86 cm*

e) According to
Section 12-2, what is the period of this oscillation?

*T
= 2pi sqrt ( m/k)*

* T
= 2 * 3.14159* sqrt (5 / 490 ) = .6347 sec*

Pg. 451 Sec Review:

1) Two mass-spring
systems vibrate with simple harmonic motion. If the spring constants of each
system are equal and the mass of one is twice that of the other, which system
has a greater period?

*The one with
twice the mass has a greater period*

* by sqrt(2) =1.41 times*

2. A child swings
on a playground swing with a 2.5 m long chain.

a)
What is the period of the child’s motion?

*for a
pendulum T = 2*pi*sqrt (l/g) *

*=
2*3.14*sqrt(2.5/9.8)*

*T=3.1735
sec/swing*

b) What is the
frequency of the vibration?

* f = 1/T = 1/3.17=.315 swings/sec=.315 Hz*

3. A pendulum
swings from maximum displacement on one side of equilibrium to maximum
displacement on the opposite side of equilibrium. If the pendulum swings
through a total of 24°, what is the amplitude of this vibration?

*Amplitude =
12 °*

Sketch a graph of
how the horizontal displacement and velocity would change over time. *Displacement = cos curve, Vel=
sin curve*

**Honors* Find the
equation that shows this.

4. The reading on
a metronome indicates the number of oscillations per minute. What are the
period and frequency of the metronome’s vibration when the metronome is set at
180?

*180
beats/min = 180 beats/60sec = 3 beats/sec = 3 Hz*

By what factor
should the length of a simple pendulum be changed if the period of vibration
were to be tripled?

A) 3 B) 1/3 ** C) 9** D)
27

A pendulum with a
mass of 0.1 kg was released. The string made an angle of 7 ° with the vertical.
The bob of the pendulum returns to its lowest point every 0.1 seconds.

What
is the period, frequency, amplitude?

*Amplitude =
7°, T = 0.2 seconds, f = 1/.2=5 Hz*

The pendulum is
replaced by one with a mass of 0.3 kg and set to swing at a 15 ° angle. How do
the period and frequency change? Why or why not?

*T and f do
not change…. Period of a pendulum is NOT affected by mass or amplitude….*

If this pendulum
was set to keep time and then brought to the moon ( gravity is 1/6^{th}
of Earth), what property would you have to change to make sure it kept the same
accurate time? *Period is affected by length, you would need to change the
length by 6 (longer)*

*** Honors**By how
much?