**Your
Answer to Q300.**
Correct answer is 20 Hz. If you did not get this answer try again.

Help: *Fundamentals of
Sound*, Secs. 2-C, 7-F.

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 300.

**Hint
for Question 300:** What is the expression for the fundamental frequency
of a string tied at both ends in terms of the wave velocity and the length of
the string?

Return to Question 300.

**Your
Answer to Q300. **Congratulations, your answer is **correct**.

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

Return to Question 300.

**Correct
Answer to Question 300:** The fundamental frequency of a standing wave
on a such a string is *v*/2*L*. So if *L* goes to *L*/2
the fundamental frequency goes up by a factor of 2 and becomes 20 Hz.

Return to Question 300.

**Your
Answer to Q310.** Sorry, your answer is **not** correct. First find
the new fundamental frequency.

Help: *Fundamentals of
Sound*, Sec. 7-F.

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 310.

**Hint
for Question 310:** If you find the fundamental frequency for the new
length, then the higher harmonics are simple multiples of that. But how are
things changed with a wave of that symmetry?

Return to Question 310.

**Your
Answer to Q310. **Congratulations, your answer is **correct**.

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

Return to Question 310.

**Correct
Answer to Question 310:** The new fundamental frequency is just twice
the old one. But the wave shown is antisymmetric, so it uses only the even harmonics
2*f*_{0} , 4*f*_{0}, etc. The old fundamental was then 5Hz and the new one
is just twice that with a string that is half as long (as in Question 300) or
10 Hz. The new second harmonic is thus 20 Hz; the new fourth harmonic 40 Hz,
etc. So the correct spectrum is b)

Return to Question 310.

**Your
Answer to Q320.** Sorry, your answer is **not** correct. What
are the requirements for a continuous spectrum?

Help: *Fundamentals of
Sound*, Sec. 7-F.

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 320.

**Hint
for Question 320:** To have a continuous spectrum the wave must be non-repeating.

Return to Question 320.

**Your
Answer to Q320. **Congratulations, your answer is **correct**.

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

Return to Question 320.

**Correct
Answer to Question 320:** The waves A) and D) are continously repeating
or periodic. Thus they have discrete spectra. Waves B) and C) are impulsive,
that is they last only for a short while and so they are the ones that have
continuous spectra. The correct answer is d)

Return to Question 320.

**Your
Answer to Q330.** Sorry, your answer is **not** correct. Does
the wave require a continuous or a discrete spectrum?

Help: *Fundamentals of
Sound*, Sec. 7-F.

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 330.

**Hint
for Question 330:** Does the wave require a continuous or a discrete
spectrum? What is the repeat frequency of the wave? Since you do not see the
wave as a function of position, symmetry is **not** a factor here.

Return to Question 330.

**Your
Answer to Q330. **Congratulations, your answer is **correct**.

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

Return to Question 330.

**Correct
Answer to Question 330:** The wave is repeating and thus requires a
discrete spectrum. Answers c) is continuous and so is eliminated. The repeat
time is 2 s, so the repeat frequency is 1/2 Hz. Only b) has a spectrum that
has a repeat frequency of 1/2 Hz, with its fundamental of 1/2 Hz. (Note symmetry
does not play a role here because the wave is shown as a function of *time*,
not position. You are seeing the motion of just a single point in space. All
waves that have the correct repeat frequencies (which you get from the time
graph) will have the appropriate symmetry in space.)

Return to Question 330.

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