Physics 110 Homework Solutions Week #10
Wednesday, November 13, 2013
Chapter 10
Questions
10.13 If the length of the string doubles, the velocity increases by 2 . As doubles then
the velocity decreases by 2 . If the tension force decreases then the velocity
decreases by 2 .
Multiple-Choice
10.1 B
10.3 A €
10.5 D
€
€
Problems
10.7 Traveling waves
a. We calculate the speed using the equation
b. We know that
.
.
10.11 Transverse waves on a string
c. Assuming that the weight of the string extending from the pulley is negligible we
have the wave speed given by
.
d. The fundamental frequency occurs when n = 1, and using
.
e. The wavelength of the fourth harmonic can be found using
.
10.15 We know that in a string of length L,
, and the distance between nodes,
, is 0.25m, therefore λ = 0.50m. Further we have
.
10.17 We must first convert the feet units to meters, where 1m = 3.28ft. If the total
length of the strings was 104ft or 31.7m, then each individual string measures
7.93m in length. Using
. For the fundamental
frequency, we know that n = 1, so that in the string of length
,
where L is 14ft or 4.27m for the entire height of the double bass, and
.
Friday, November 15, 2013
Chapter 11
Questions
11.3 Δβ = 10log(I2/Io) – 10log(I1/Io) = 10 log(I2/I1) so 100.1 = I2/I1 = 1.25
similarly a factor of 2: β = 10log2 = 3.0 dB
Multiple-Choice
11.2 C
11.3 B
11.7 C
11.8 D
Problems
11.1 A swimming beaver
a. First, we must convert the distance between shores from 800ft to metric units.
We know that
, and that the velocity of sound in air is 343m/s.
Therefore, it takes tair = 0.71s for the sound of the slap above the water surface to
cross the lake.
b. Using
, and vsound,water = 1482m/s, we find that it takes twater =
0.16s for the sound of the slap under water to cross the lake.
11.12 A crying baby
a. A sound intensity of 8x10-6 W/m2 corresponds to
dB.
b. If two infants are crying then the intensity is twice the intensity of one of the
children, assuming that they are crying at the same intensity. Thus we have for
the intensity level
dB.
c. The general formula for any even number of crying children is
.
d. The time is given by
.
11.13 We know that it takes 5 seconds for the sound to propagate over the unknown
distance. Thus we have
.
11.23 Oscillating String
c. We know that
and for the second harmonic
. We can determine
from the given information that
. Now using
.
d. For the second harmonic we have 2 oscillating lobes, therefore doubling the
frequency means that we are observing the fourth harmonic, where n = 4.
Therefore, the number of oscillating lobes is 4.
e. Since
for a tube with both ends open, we find that L = 0.52m.
f. The frequency and the length are inversely related. For a longer length of pipe we
have a lower frequency. Thus the beat frequency is 325 Hz and from this we
calculate the length of pipe that we have. We find
. The pipe from part c is 0.520m long
so we need to shorten this pipe by 0.008m or by
.
g. These frequencies represent odd multiples of the fundamental frequency and thus
the pipe is closed.