**Answer
to Q145**: The answer is 1 s. If you did not get that correct, perhaps you
should realize that you need to find the period of the wave. How would
you do that?

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

Would you like a HINT?

You should really try to reason this on your own, but if you want to read it, the correct answer is HERE.

Return to Question
145.

**Hint
for Q145**: You need to find the period. To do that you need the wavelength,
from which you can get the frequency, and then the period.

Return
to Question 145.

**Correct
answer to Q145: **The form shown occurs 1/2 cycle after the top form.
So the time needed is 1/2 period. To find the period we need *f = v/*.
To find ,
we need to figure out what
fraction of the total length of 14 ft is a wavelength. One way to do this is
to divide the wave up into 7 parts: each one quarter of a wavelength, that is,
from the wall on the left to the first peak, from the peak to the next zero,
etc. There are 7 such parts in the total wave, each 2 ft long. A wavelength
is 4 of these so the wavelength is 8 ft. The frequency is (4ft/s)/(8ft) = 0.5Hz.
Thus the period is *T=*1*/f *= 2 s. and each cycle takes 2 s. The
position we want is at a half cycle of the motion so it takes only 1 s to get
there. The answer is 1 s.

Return
to Question 145.

**Your
answer to Q147**:
Sorry, your answer is **not **correct. Think what pressure you must
have at an open end.

Help: *Fundamentals of
Sound*, Sec. 3-A.

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
147.

**Hint
for Q147**: The outside atmosphere is at normal pressure; what does
that say about the pressure at an open end? For a closed end you should think
what happens to the pressure as the molecules move toward or away from that
open end.

Return
to Question 147.

**Your
answer to Q147: **Congratulations, your answer is **correct**!

To see the "official" correct answer click HERE.

Return
to Question 147.

**Correct
answer to Q147**: Since the open end is open to the entire atmosphere,
which is at normal pressure, the pressure there stays **normal**. Any compression
or rarefaction there is relieved immediately by the surrounding atmosphere.
At a closed end, if the molecules are moving away from that end, there are none
to replace them so that a rarefaction forms there. If the molecules are moving
toward the closed end, they run smack into the end and crowd together, like
passengers all moving to the rear of the bus, forming a compression. There is
always a pressure antinode at a closed end. Thus in the figure there is an open
end at the left and a closed end at the right, and c) is correct.

Return
to Question 147.

**Your
answer to Q148**: Sorry, your answer is **not** correct.

Help: *Fundamentals of
Sound*, Secs. 3-A, 3-B, 3-D.

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
148.

**Hint
for Q148**: Displacement waves have properties opposite to pressure
waves: where one has nodes the other has antinodes and vice versa. That eliminates
two of the four answers. Can you figure out how to tell which of the remaining
answers is the correct one?

Return
to Question 148.

**Your
answer to Q148**: Congratulations, your answer is **correct**!

To read the "official" correct answer click HERE.

Return
to Question 148.

**Correct
answer to Q148**: At the open end of a tube, the pressure is kept normal
by molecules streaming in and out relieving any compression or rarefaction that
tries to form there. Thus there is a displacement antinode at an open end. Molecules
at a closed end are stuck there (the forces of the wall hold them tight), and
so at a closed end there is a displacement node. Thus the answer must be either
c) or d). In the figure there is a rarefaction a bit in from the the open left
end. Molecules are of course streaming away from a rarefaction—that is
why the density and pressure are lower there. Thus at the very left end molecules
must be moving to the left out of the open end of the tube. But a leftward
displacement is negative (by definition). This immediately says the answer is
d).

Further along in the tube there is a compression—where the pressure reaches a positive peak. The molecules must be moving toward a compression—that is why there is crowding there. That means they are moving to the right (positive displacement) on the left of the compression, and to the left (negative displacement) on the right of the compression. This is exactly what is happening in d).

Return
to Question 148.

**Your
answer to Q150**: Sorry, your answer is **not **correct. What
is the period of the wave?

Help: *Fundamentals of
Sound*, Secs. 2-D, 3-D.

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
150.

**Hint
for Q150**: You found the period in Questions 145. It was 2 s. How many
cycles will be completed in 9.5 s?

Return
to Question 150.

**Your
answer to Q150**: Congratulations, your answer is **correct**!

To read the "official" correct answer, click HERE.

Return to Question
150.

**Correct
answer to Q150**: The period is 2 s, so the wave completes 4 of these—returning
to the original form—in 8 s. There remains 1.5 s, and in 1 s the wave will
be as shown in c). However in 0.5 s left over it moves half way back to the
original position, so it is as in a) after 9.5 s. The answer is a).

Return
to Question 150.

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