1. PHY 231 1 PHYSICS 231 Lecture 38: Resonances, beats and review Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom

2. PHY 231 2 standing waves in a rope both ends fixed n =2L/n or L=n n /2 F: tension in rope  : mass per unit length f 1 : fundamental frequency

3. PHY 231 3 Both ends open

4. PHY 231 4 One end open, one end closed even harmonics are missing!!!

5. PHY 231 5 example A simple flute is played by blowing air in on one side and the other end is open. The length of the tube can be varied manually (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 change v=343m/s? f 1 =343/(4*0.5)=172 Hz f 3 =3*343/(4*0.5)=514 Hz f 1 =343/(4*0.25)=343 Hz f 3 =3*343/(4*0.25)=1028 Hz

6. PHY 231 6 example A simple flute is played by blowing air in on one side and the other end is closed. The length of the tube can be varied manually (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 change v=343m/s? f 1 =343/(2*0.5)=343 Hz f 2 =2*343/(2*0.5)=686 Hz f 1 =343/(2*0.25)=686Hz f 2 =2*343/(2*0.25)=1372 Hz

7. PHY 231 7 harmonics Generally speaking, many harmonics with different intensities can be present at the same time. L +

8. PHY 231 8 beats Superposition of 2 waves with slightly different frequency The amplitude changes as a function of time, so the intensity of sound changes as a function of time. The beat frequency (number of intensity maxima/minima per second): f beat =|f a -f b | DEMO

9. PHY 231 9 example Someone is trying to tune a guitar. One of the strings is supposed to have a frequency of 500 Hz. The person is using a tuning fork which produces a sound of exactly this frequency, but while sounding the fork and the playing the guitar, hears a beat in the sound with a frequency of 3 Hz (3 beat per second). a) What is the real frequency of the guitar string? b) By what fraction does the person need to change the tension of the guitar string to tune it properly? a) f b =|f fork -f guitar | 3=|500-f guitar | f guitar =497 or 503 Hz b) so f~  F f current /f ideal =  (F current /F ideal ) 497/500=0.954 or 503/500=1.006 F ideal =F current /(0.994) 2 =1.012F current or F ideal =F current /(1.006) 2 =0.988F current

10. PHY 231 10 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 natural frequency: large oscillations occur: Resonance demo Resonances occur in many daily situations: shock absorber in car playing basketball resonating lecture room!! Famous example: Tacoma bridge

11. PHY 231 11 see also recitations on Thursday (12-1 and 17:15-18:15) review on Friday any more??? today: one problem each from ch. 2,3,4,5,6 Friday: one problem each from 7,8,9 and the rest 10,11,12,13,14 (2 each) Start with the sample problems on the web to see how you stand on each chapter!!

12. PHY 231 12 chapter 2. A person throws 2 stones from the top of a building with a speed of 20 m/s. One is thrown up, and the other is thrown down. The first one hits the street after 5 s. How much later does the second one hit? Stone thrown down: x(t)=x(0)+v(0)t+½at 2 =h-20t-½(9.8)t 2 if t=5, x=0, so 0=h-100-½(9.8)5 2 so h=222.5 m Stone thrown up: x(t)=x(0)+v(0)t+½at 2 =h+20t-½(9.8)t 2 =222.5+20t-½(9.8)t 2 when it reaches the ground, x=0 so 0=222.5+20t-½(9.8)t 2 so: -4.9t 2 +20t+222.5=0 you’ll find t=-5 or t=9.1 s must be 9.1 s difference between the times that the stones hit: 9.1-5=4.1 s x(t)=x(0)+v(0)t+½at 2 v(t)=v(0)+at

13. PHY 231 13 chapter 3 v0v0 30m 3m A car is trying to jump over a 30m-wide river using a ramp of 3 m high set at an angle of 30 0 with the horizontal. a) What is the minimum velocity v 0 required to cross the bridge? b) What is the highest point of the car?  =30 0 a)x(t)=x(0)+v x (0)t+½at 2 =v 0 tcos  =0.866v 0 t must at least be 30m, so 30=0.866v 0 t and thus t=34.6/v 0 y(t)=y(0)+v y (0)t-½gt 2 =y(0)+v 0 tsin  -4.9t 2 =3+0.5v 0 t-4.9t 2 when it hits the ground: y(t)=0=3+0.5v 0 t-4.9t 2 use t=34.6/v 0 and find: 0=3+0.5*34.6-4.9(34.6/v 0 ) 2 solve for v 0 and find v 0 =17 m/s (61.2 km/h) b)At highest point: vertical component of velocity=0 v y (t)=v y (0)+at=v 0 sin  -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