**Your
Answer to Q550.**
Sorry, your answer is **not** correct. Use the Fletcher-Munson graph.

Help: *Fundamentals of
Sound*, Sec. 11-I.

Or would you like a **HINT**?

You should try to work out the answer on your own, but, if you insist on reading it, the correct answer is here.

Return to Question 550.

**Hint
for Question 550:** Use the Fletcher-Munson graph. Follow the 10 phon
contour.

Return to Question 550.

**Your
Answer to Q550. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 550.

**Correct
Answer to Question 550:** To answer the question, look at the 10 phon contour
and find where it meets the vertical line corresponding to 100 Hz; you now read
the decibel level off the left axis to be about 45 dB, answer c).

Return to Question 550.

**Your
Answer to Q560.** Sorry, your answer is **not** correct. Use
the Fletcher-Munson graph.

Help: *Fundamentals of
Sound*, Sec. 11-I.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here.

Return to Question 560.

**Hint
for Question 560:** Use the Fletcher-Munson graph. Find the phon contour corresponding
to each frequency and intensity.

Return to Question 560.

**Your
Answer to Q560. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 560.

**Correct
Answer to Question 560:** From the Fletcher-Munson diagram, find the 60 dB
intensity level on the left-most vertical axis and look horizontally across
to where the 100 Hz vertical line crosses it. You can see that this point is
midway between the 30 and 40 phon contours, so let's approximate the loudness
level of this sound as 35 phons. Now find the 60 dB intensity level on the left
and trace it across to where the 10,000 Hz vertical line crosses it. You can
see that this point lies between 40 and 50 phons, but that it is much closer
to the 50 phon contour. So let's approximate its loudness level as 48 phons.
Therefore, because the loudness level (48 phons) of the 10,000 Hz tone at an
intensity of 60 dB has a larger value than the loudness level (35 phons) of
the 100 Hz tone played at same intensity level, the 10,000 Hz tone sounds louder,
answer b).

Return to Question 560.

**Your
Answer to Q570.** Sorry, your answer is **not** correct. Use the
Fletcher-Munson graph.

Help: *Fundamentals of
Sound*, Sec. 11-I.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here.

Return to Question 570.

**Hint
for Question 570:** Use the Fletcher-Munson graph. First find the phon
level for the 10,000 Hz tone.

Return to Question 570.

**Your
Answer to 570. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 570.

**Correct
Answer to Question 570:** Find the 90 dB intensity level on the left
and trace it across to where it intersects the 10,000 Hz vertical line. Notice
that you are almost on top of the 80 phon contour. To have the same loudness,
the 500 Hz tone must also have a loudness level of 80 phons. So follow the 80
phon contour to 500 Hz, and read across to find the intensity level of 80 dB.

Return to Question 570.

**Your
Answer to Q600.** Sorry, your answer is **not** correct. You have
to convert to an intensity before you multiply by 3.

Help: *Fundamentals of
Sound*, Sec. 11-K.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer it is here.

Return to Question 600.

**Hint
for Question 600:** Convert to intensity ratio of *I*/*I*_{min}
and then multiply by 3.

Return to Question 600.

**Your
Answer to Q600. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 600.

**Correct
Answer to Question 600:** For 85 dB we have *I*/*I*_{min}
= 10^{8.5}. Multiply this by 3, which from the log table is close to
10^{0.5} so that 3x*I*/*I*_{min} = 10^{0.5+8.5}
= 10^{9.0}. The exponent is the number of Bels, so the number of decibels
is 90. c) is the answer.

Return to Question 600.

**Your Answer to Q610.** Sorry, your answer is
**not** correct. One of the intensities is much bigger than the others. How
do you tell that? Why does that help?

Help: *Fundamentals of
Sound*, Sec. 11-K.

Or would you like a **HINT**?

You should try to work out the answer on your own, but if you insist on reading it the correct answer is here.

Return to Question 610.

**Hint
for Question 610:** One of the intensities is much bigger than the others
because the decibel level is related to the log of the intensity.

Return to Question 610.

**Your
Answer to Q610. **Congratulations, your answer is **correct**.

If you would like, you can compare your answer to the "official" correct answer

Return to Question 610.

**Correct
Answer to Question 610:** b) is correct. The 67 dB sound is much more
intense than the other two. To prove this we *will* do some math: To make
an easier calculation than the one in the problem consider 70 dB and 50 dB.
The associated total intensity ratio is *I*/*I*_{min} = 10^{7}
+ 10^{5} = 1.0 x 10^{7} + 0.01 x 10^{7} = 1.01 x 10^{7}
= 10^{7.004}. The sound level is hardly different from 70 dB. The same
is true in the real problem; the 67 dB sound is much more intense than the 43
dB or even the 50 dB, and the total intensity is still closest to 67 dB. If
you do the math carefully you find that the exact answer is 67.10 dB. But the
math was not necessary to see the answer to the problem.

Return to Question 610.

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