PHY 2311

PHYSICS 231Lecture 38: Resonances, beats and

review

Remco ZegersQuestion hours: Thursday 12:00-13:00 & 17:15-

18:15Helproom

PHY 2312

standing waves in a rope

both ends fixed n=2L/n or L=nn/2

F

L

n

L

nvvf

nn 22

F: tension in rope: mass per unit length

1

2

1

2

2

22

nfL

nvf

L

vf

L

vf

n

f1: fundamental frequency

PHY 2313

Both ends open

3,2,12 1 nnfL

nvfn

PHY 2314

One end open, one end closed

5,3,14 1 nnfL

nvfn

even harmonics are missing!!!

PHY 2315

example

A simple flute is played by blowing air in on one side and theother end is open. The length of the tube can be variedmanually (like a trombone). What are the frequencies of the first two possible harmonics if L=0.5m? If the length is made half of the original length, how will these changev=343m/s?

f1=343/(4*0.5)=172 Hz f3=3*343/(4*0.5)=514 Hz

f1=343/(4*0.25)=343 Hz f3=3*343/(4*0.25)=1028 Hz

PHY 2316

example

A simple flute is played by blowing air in on one side and theother end is closed. The length of the tube can be variedmanually (like a trombone). What are the frequencies of the first two possible harmonics if L=0.5m? If the length is made half of the original length, how will these changev=343m/s?

f1=343/(2*0.5)=343 Hz f2=2*343/(2*0.5)=686 Hz

f1=343/(2*0.25)=686Hz f2=2*343/(2*0.25)=1372 Hz

PHY 2317

harmonics

Generally speaking, many harmonics with different intensitiescan be present at the same time.

L

+

PHY 2318

beatsSuperposition of 2 waves with slightly different frequency

The amplitude changes as a function of time, so the intensityof sound changes as a function of time. The beat frequency (number of intensity maxima/minima per second): fbeat=|fa-fb|

DEMO

PHY 2319

example

Someone is trying to tune a guitar. One of the strings issupposed to have a frequency of 500 Hz. The person isusing a tuning fork which produces a sound of exactlythis frequency, but while sounding the fork and the playingthe guitar, hears a beat in the sound with a frequency of3 Hz (3 beat per second). a) What is the real frequency ofthe guitar string? b) By what fraction does the person need tochange the tension of the guitar string to tune it properly?

a) fb=|ffork-fguitar| 3=|500-fguitar| fguitar=497 or 503 Hz

b)

F

L

nfn 2

so f~F

fcurrent/fideal= (Fcurrent/Fideal)497/500=0.954 or 503/500=1.006Fideal=Fcurrent/(0.994)2=1.012Fcurrent

or Fideal=Fcurrent/(1.006)2=0.988Fcurrent

PHY 23110

Resonances

Realistically, oscillations are damped due to frictional forces.However, we can drive the oscillation via an external source.Example: mass on a spring: natural frequency f=1/(2)(k/m)

If the frequency of the driving force equals the naturalfrequency: large oscillations occur: Resonance demo

Resonances occur in many daily situations:•shock absorber in car•playing basketball•resonating lecture room!!

Famous example: Tacoma bridge

PHY 23111

see also recitations on Thursday (12-1 and 17:15-18:15)review on Fridayany more???

today: one problem each from ch. 2,3,4,5,6Friday: one problem each from 7,8,9 and the rest 10,11,12,13,14

(2 each)

Start with the sampleproblems on the web tosee how you stand oneach chapter!!

PHY 23112

chapter 2.A person throws 2 stones from the top of a building with aspeed of 20 m/s. One is thrown up, and the other is throwndown. The first one hits the street after 5 s. How much laterdoes the second one hit?

Stone thrown down:x(t)=x(0)+v(0)t+½at2=h-20t-½(9.8)t2

if t=5, x=0, so 0=h-100-½(9.8)52 so h=222.5 m

Stone thrown up:x(t)=x(0)+v(0)t+½at2=h+20t-½(9.8)t2=222.5+20t-½(9.8)t2

when it reaches the ground, x=0 so0=222.5+20t-½(9.8)t2 so: -4.9t2+20t+222.5=0you’ll find t=-5 or t=9.1 s must be 9.1 sdifference between the times that the stones hit: 9.1-5=4.1 s

x(t)=x(0)+v(0)t+½at2

v(t)=v(0)+at

PHY 23113

chapter 3v0

30m3m

A car is trying to jump over a 30m-wide river using a ramp of3 m high set at an angle of 300

with the horizontal. a) What is theminimum velocity v0 required to cross the bridge? b) What is thehighest point of the car?

=300

a) x(t)=x(0)+vx(0)t+½at2=v0tcos=0.866v0t must at least be 30m, so 30=0.866v0t and thus t=34.6/v0

y(t)=y(0)+vy(0)t-½gt2=y(0)+v0tsin-4.9t2=3+0.5v0t-4.9t2

when it hits the ground: y(t)=0=3+0.5v0t-4.9t2

use t=34.6/v0 and find: 0=3+0.5*34.6-4.9(34.6/v0)2

solve for v0 and find v0=17 m/s (61.2 km/h)b) At highest point: vertical component of velocity=0 vy(t)=vy(0)+at=v0sin-9.8t=17*0.5-9.8t=0 t=0.87 s y(0.87)=3+0.5*17*0.87-4.9(0.87)2=6.7 m