**Your
answer to Q130: **Sorry,
your answer is **not **correct. The string moves one cycle in a period.

Help: *Fundamentals of
Sound*, Sec. 2-C.

Or, would you like a HINT?

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

Return to Question
130.

**Hint
for Q130**: The standing wave will move one cycle, that is back to the
original picture, in one period; and one/half cycle in one/half period. You
need to figure out what a period is. You can get this from knowing the frequency.
How do you find that?

Return to Question
130.

**Your
answer to Q130**: Congratulations, your answer is **correct**!

To see the "official" correct answer click HERE.

Return to Question
130.

**Correct
answer to Q130**: The frequency is just the wave velocity divided by
the wavelength. Since the string is 12 ft long, the wavelength is just 2/3 of
this or 8 ft. From this we find the frequency is *v/*l
= (16 ft/s)/8 ft = 2 Hz. And the period = 1/*f *= 0.5 s.
Since 1 s is two periods and 0.25 s is 1/2 period, the string will have moved
2.5 complete cycles. Two cycles gets it back to where it started and the extra
half cycle gets it to the position shown in c), which is the correct answer.

Return to Question
130.

**Your
answer to Q135**: Sorry, your answer is **not **correct. What
portion of a period is required to get to the position shown?

Help: *Fundamentals of
Sound*, Sec. 2-C.

Or, would you like a HINT?

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

Return to Question
135.

**Hint
for Q135**: The position shown is just one-fourth of a cycle away from
the original position. How long does it take to get there?

Return to Question
135.

**Your
answer to Q135**: Congratulations, your answer is **correct**!

To see the "official" correct answer click HERE.

Return to Question
135.

**Correct
answer to Q135**: The position shown is just one-fourth of a cycle away
from the original position. In the next quarter cycle the string would have
moved to the position shown in answer c) of the previous question and so on.
The period here is just as in the last question, 0.5 s, so what we need is one-fourth
of 0.5 s or (1/4)x(1/2s)=1/8 s.

Return to Question
135.

**Your
answer to Q137**: Sorry, your answer is **not **correct. What
is the frequency of the fundamental normal mode on the violin string?

Help: *Fundamentals of
Sound*, Sec. 2-C.

Or, would you like a HINT?

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

Return to Question
137.

**Hint
for Q137**: The frequency of the fundamental normal mode on the violin
string is *f*_{0} = *v/(2L)*. What does that tell you to do
to increase the frequency?

Return
to Question 137.

**Your
answer to Q137**: Congratulations, your answer is **correct**!

To see the "official" correct answer click HERE.

Return to Question
137.

**Correct
answer to Q137**: The frequency of the fundamental normal mode on the
violin string is *f*_{0} = *v/(2L)*. So if you want to increase*
f* you must increase *v* or decrease *L*. Those are the choices
in a) and b). Moving the bow faster just makes the same sound of the same frequency
more often or perhaps continuously. d) is the correct answer.

Return
to Question 137.

**Your
answer to Q140**: Sorry, your answer is **not **correct. Did you
try counting the number of nodes or antinodes?

Help: *Fundamentals of
Sound*, Sec. 2-D.

Or, would you like a HINT?

You should really try to work out the answer on your own, but of you insist on readng it, the correct answer is HERE.

Return to Question
140.

**Hint
for Q140**: The fundamental of a string tied at one end and loose at
the other has one antinode and one node. The first overtone has two antinodes
and two nodes, etc. Be careful with the harmonic numbering, however!

Return to Question
140.

**Your
answer to Q140**: Congratulations, your answer is **correct**!

To see the "official" correct answer click HERE.

Return to Question
140.

**Correct
answer to Q140**: The fundamental of a string tied at one end and loose
at the other has one antinode and one node (at the tied end). The first overtone
has two antinodes and two nodes; the second overtone has three antinodes and
three nodes, etc. However, the first overtone is the **third** harmonic,
and the second overtone is the fifth harmonic, etc, because the odd harmonics
don't exist with these boundary conditions. Since this mode has 4 antinodes
and 4 nodes, it is the third overtone and the 7th harmonic, just by enumeration.
See Fig. 2-10 in the *Fundamentals of Sound*. The correct answer is then
e) none of the above.

Return
to Question 140.

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