A. PH 105-003/4 ----Monday, Nov. 19, 2007
Homework:
PS 13, Chapter 15, is due Wed. at 11PM
Clicker question feedback:
see WebAssign Forum
Re-do clicker question on scientific notation
Chapter 16: we’ll skip Sec. 16.4-5-6
(reflection, energy transfer, wave equation)
Chapter 17: Sound waves: Active Figure 17.2,
v2 = B/r
A.
B.
C.
D.
A. PH 105-003/4 ---- Monday, Nov. 26, 2007
Homework:
PS 14, Chapter 16 (short), due Wed. at
11PM
Clicker question feedback: 4 responses
Problem Session: Weds. 5 PM (Jones
Th@5?)
E. Exam Friday:
F. Ch. 13.3 (Kepler) – Ch. 19.5 (ideal gas)
G. [may skip 19.4, thermal expansion]
A. PH 105-003/4 ---- Monday, Nov. 26, 2007
Chapter 17:
I = intensity = power/area
10-12 w/m2 = ideal threshold of hearing = I0
10-7 w/m2 = conversation
use log scale
“sound level” b = 10 log10 I/I0
(units decibels)
e.g., I/I0 = 100,000 ↔ b = 50 decibels
Clicker question (Nov 19): According to R. Serway,
a mosquito buzz has intensity 10-8 w/m2. How many
decibels is this?
40
1
Threshold
of hearing
depends on
frequency –
but is about
10-12 watts for
a range of
frequencies
(50 Hz –
12KHz).
Free www
hearing test:
w/ left aid
on high
PH 105-003/4 ---- Monday, Nov. 26, 2007
Chapter 17: Point source:
Total power P = area*I
P = 4p r2 I
I=?
r
2nd Clicker question (Nov 25): If the sound of a
mosquito has intensity 10-8 w/m2 at a distance
of 0.2 m, what is the total sound power
produced by the mosquito, in nanowatts?
5
0.5
A.
PH 105-003/4 ---- Monday, Nov. 26, 2007
Chapter 17: Doppler Effect
moving source and/or observer
Active Figure 17.8
Apparent frequency (heard by observer):
v  vobserver
f '
f
v  vsource
Clicker Question: While you are traveling
at 85 mph, a state trooper driving behind
you (moving in the same direction) at 85
mph turns on his 500 Hz siren. The
frequency you hear is
A. Less than 500 Hz
B. Exactly 500 Hz
C. More than 500 Hz
A.
PH 105-003/4 ---- Monday, Nov. 26, 2007
Solution: Plug into equation:
vs = +85 mph (toward observer)
vo = -85 mph (away from source)
v  vobserver
v  (85 mph )
f '
f 
f  f  500 Hz
v  vsource
v  (85 mph )
A.
PH 105-003/4 ---- Monday, Nov. 26, 2007
Solution: Plug into equation:
vs = +85 mph (toward observer)
vo = -85 mph (away from source)
v  vobserver
v  (85 mph )
f '
f 
f  f  500 Hz
v  vsource
v  (85 mph )
A.
PH 105-003/4 ---- Monday, Nov. 26, 2007
Chapter 18: Standing waves.
Review: On a string of length L, there can
be n half-wavelengths: L = n l/2, so ln = 2L/n
Resulting frequencies: fn = v/ln  n v/2L = n f1
In an air column,
closed end  displacement node
2 closed ends: like string, f = f1, 2f1, 3f1, ...
open end  pressure node 
displacement antinode
One open, one closed: L = n l/2 + l/4
f = f0, 3f0, 5f0, …
Sec. 18.6: 1D vs. 2D vibrations
[or, why 1D objects (strings, horns) make music,
while 2D objects (drumheads) make noise]
Vibrations of 1D objects: fn = n f1 (integer ratios)
Normal modes of a drumhead (Fig. 18.16)
½ # azimuthal nodes, # radial nodes
frequency f
A. PH 105-003/4 ---- Monday, Dec. 3, 2007
Exam back
Homework due Friday
Closing assessment Wednesday
Chapter 18.5: Beats
Active Figure 18.17
fbeat = f2 – f1
Chapter 19: Temperature
[Anonymous] Clicker question:
It has been suggested that we have a “lab review”
or “lab final” instead of the last lab this Thursday,
in which each individual student would do one or
two pieces of a previous lab. Rate this as a
A.
B.
C.
D.
E.
Very good idea
Good idea
No opinion
Bad idea
Very bad idea