Physics 207: Lecture 30, Pg 1 Lecture 30Goals: Wrap up chapter 20. Wrap up chapter 20. Review for the final. Review for the final. Final exam on Friday, Dec 23, at 10:05 am in BASCOM 272. Final will cover Chapters 1-20 (excluding Chapter 13). Semi-cumulative. HW 12 due tomorrow night.

Physics 207: Lecture 30, Pg 2 The Doppler effect l The frequency of the wave that is observed depends on the relative speed between the observer and the source. vsvs observer

Physics 207: Lecture 30, Pg 3 The Doppler effect f=f 0 /(1-v s /v) l Approaching source: l Receding source: f=f 0 /(1+v s /v)

Physics 207: Lecture 30, Pg 4 l The figure shows a snapshot graph D(x, t = 2 s) taken at t = 2 s of a pulse traveling to the left along a string at a speed of 2.0 m/s. Draw the history graph D(x = −2 m, t) of the wave at the position x = −2 m. Waves

Physics 207: Lecture 30, Pg 5 l History Graph: time (sec) D(x = −2 m, t)

Physics 207: Lecture 30, Pg 6 l A concert loudspeaker emits 35 W of sound power. A small microphone with an area of 1 cm 2 is 50 m away from the speaker. l What is the sound intensity at the position of the microphone? l How much sound energy impinges on the microphone each second?

Physics 207: Lecture 30, Pg 7 R P source =35 W R=50 m l The power hitting the microphone is: P microphone = I A microphone

Physics 207: Lecture 30, Pg 8 Intensity of sounds l If we were asked to calculate the intensity level in decibels: I 0 : threshold of human hearing I 0 = W/m 2

Physics 207: Lecture 30, Pg 9 l Suppose that we measure intensity of a sound wave at two places and found them to be different by 3 dB. By which factor, do the intensities differ?

Physics 207: Lecture 30, Pg 10 Engines l For the engine shown below, find, W out, Q H and the thermal efficiency. Assume ideal monatomic gas. PiPi 4P i ViVi 2V i Q=90J Q=25J

Physics 207: Lecture 30, Pg 11 l First, use the ideal gas law to find temperatures PiPi 4P i ViVi 2V i Q=90J Q=25J TiTi 4T i 8T i 2T i l From the right branch, we have: nC V ΔT=90 J n(3R/2)6T i =90J nRT i =10J

Physics 207: Lecture 30, Pg 12 l Work output is the area enclosed by the curve: PiPi 4P i ViVi 2V i Q=90J Q=25J TiTi 4T i 8T i 2T i W out =area=3P i V i =3nRT i =30J

Physics 207: Lecture 30, Pg 13 l From energy conservation: W out =Q H -Q C W out =30J Q C =115J η=0.2 l The thermal efficiency is: Q H =145J

Physics 207: Lecture 30, Pg 14 The Carnot Engine l All real engines are less efficient than the Carnot engine because they operate irreversibly due to the path and friction as they complete a cycle in a brief time period. l Carnot showed that the thermal efficiency of a Carnot engine is:

Physics 207: Lecture 30, Pg 15 l For which reservoir temperatures would you expect to construct a more efficient engine? A) T cold =10 o C, T hot =20 o C B) T cold =10 o C, T hot =800 o C C) T cold =750 o C, T hot =800 o C

Physics 207: Lecture 30, Pg 16 Kinetic theory l A monatomic gas is compressed isothermally to 1/8 of its original volume. l Do each of the following quantities change? If so, does the quantity increase or decrease, and by what factor? If not, why not? a. The temperature b.The rms speed v rms c. The mean free path d. The molar heat capacity C V

Physics 207: Lecture 30, Pg 17 N/V: particles per unit volume l Mean free path is the average distance particle moves between collisions: l The average translational kinetic energy is: ε avg =(1/2) mv rms 2 =(3/2) k B T C V =3R/2 (monatomic gas) l The specific heat for a monatomic gas is:

Physics 207: Lecture 30, Pg 18 Simple Harmonic Motion l A Hooke’s Law spring is on a horizontal frictionless surface is stretched 2.0 m from its equilibrium position. An object with mass m is initially attached to the spring however, at equilibrium position a lump of clay with mass 2m is dropped onto the object. The clay sticks. What is the new amplitude? k m k m (≡X eq ) 2m

Physics 207: Lecture 30, Pg 19 ½ k A 2 =½ m v max 2 v max =ωA l The speed when the mass reaches the equilibrium position: l The speed after clay sticks can be found using momentum conservation: m v max =(m+2m)v new v new =v max /3 l The new amplitude can be found using energy conservation: ½ (m+2m)v new 2 =½ k A new 2 A new =A/√3

Physics 207: Lecture 30, Pg 20 Fluids l What happens with two fluids? Consider a U tube containing liquids of density  1 and  2 as shown: Compare the densities of the liquids: (A)  1 <  2 (B)  1 =  2 (C)  1 >  2 11 22

Physics 207: Lecture 30, Pg 21 Fluids l What happens with two fluids?? Consider a U tube containing liquids of density  1 and  2 as shown: At the red arrow the pressure must be the same on either side.  1 x =  2 y  Compare the densities of the liquids: (A)  1 <  2 (B)  1 =  2 (C)  1 >  2 11 22 x y