Tutorial
Answers 85-100

**Your
answer to Q85. **Sorry,
your answer is **not** correct. Think how might you control the values of
any of these three quantities?

Help: *Fundamentals of
Sound*, Secs. 1-G, 1-H.

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
85

**Hint
for question 85. **Think how you might change the value of frequency,
wavelength, or wave velocity with a **given ** medium. If you must change
the medium to change the size of X, then that eliminates being able to create
a wave of arbitrary X.

Return to Question 85

**Your
answer to Q85. **Congratulations, your answer is correct!

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

Return to Question 85

**Correct
answer to Q85: **The wave velocity depends on the properties of the medium,
especially elasticity and inertia. The more elastic the medium is, and the less
the mass density, the faster the wave travels. So, if those are "given" properties
of the medium, we cannot change them or the wave velocity. However, one can
change the frequency of the wave by changing how you make it. If I am making
a wave on a rubber rope, I can shake my hand up and down faster to raise the
frequency. If I increase the frequency, the wavelength becomes shorter. Thus
I can control wavelength as well as frequency. The answer is d).

Return to Question
85

**Your
answer to Q90. **Sorry, your answer to question 90 is **not** correct.
Think what it is that determines wave velocity.

Help: *Fundamentals of
Sound*, Secs. 1-G, 1-H.

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 90.

**Hint
for question 90. **Wave velocity is determined strictly by the properties
of the medium.

Return to Question 90.

**Your
answer to Q90. **Congratulations, your answer is **correct**!

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

Return to Question
90.

**Correct
answer to Q90. **Wave velocity is determined by elasticity and inertia.
If I stretch the string tighter or have a less dense string, the wave will move
faster. Shaking it up and down at a higher frequency does not change the wave
velocity. The correct answer is d).

Return to Question
90.

**Answer
to Q95**: The answer is 16 feet.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

If you got that wrong,would you like a HINT?

To see how to get the correct answer click HERE

Return to Question 95.

**Hint
for Q95: **Wavelength is the distance between two adjacent crests of
a wave, i.e., the peak-to-peak distance.

Return to Question 95.

**Correct
answer to Q95:** The distance from one peak to the next is 16 feet. You can
also take the distance from any one point to the next precisely similar point—for
example, from a point where the displacement is zero, such as at *x*
= 0, to the next similar zero (at *x* = 16ft) where it has completed one
cycle in space.

Return to Question
95.

**Answer
to Q96: **The correct answer is 8 seconds.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

If you got that wrong, would you like a HINT?

To see how to reach the correct answer click HERE.

Return to Question 96.

**Hint
for Q96.** The period is the time for one complete cycle.

Return to Question 96.

**Correct
answer to Question 96:** The period is the time for one complete cycle. From
the time graph we see that the point at *x* = 16 ft starts at height 3
ft at *t* = 0, goes down to zero at* t *= 2 s, continues down to height
–3 ft at *t* = 4 s, and comes back up to height 3 ft at *t* =
8 s. The time for this complete cycle is the period and is 8 seconds.

Return
to Question 96.

** **

**Answer
to Q97**: The correct answer is 3 feet.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

If you got that wrong, would you like a HINT.

To see how to reach the correct answer click HERE.

Return to Question 97.

**Hint
for Q97**: Amplitude is the distance from the zero position (at-rest
or equilibrium position) of any point to the point where is at a maximum distance
away from equilirium.

Return to Question 97.

**Correct
answer to Q97**: The amplitude is 3 feet. This is the distance from the at-rest
position to the position of a peak of a wave. You can use either graph to find
this.

Return to Question 97.

**Answer
to Q98**: The correct answer is 2 ft/s.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

If you got that wrong, would you like a HINT?

To see how to reach the correct answer click HERE.

Return to Question 98.

**Hint
for Q98**: Wave velocity is the distance traveled divided by the time
to travel that distance. Any wave travels one wavelength in one period.

Return to Question 98.

**Correct
answer to Q98**: The wave travels one wavelength in one period. We know
that the wavelength is 16 feet and the period is 8 seconds. Thus the wave velocity
(distance over time) is 16 ft/8 s= 2 ft/s. Another way to do this is to use
the formula velocity = frequency times wavelength. The frequency is the inverse
of the period =1/8 Hz. Thus *v* = (1/8)16 = 2 ft/s.

Return to Question 98.

**Your
answer to Q99: **Sorry, your answer is **not **correct. Use
the top graph in your analysis.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

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 99.

**Your answer to Q99: **Congratulations, your
answer is **correct!**

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

Return to Question 99.

**Hint
for Q99:** Use the graph "At 2 s" to locate what the height is when the "distance
in feet" is 18 feet; this is near the middle of the graph.

Return to Question 99.

**Correct
answer to Q99: **The displacement is about 2 feet. Note that at distance (*x*)
= 16 feet, the displacement is 0 ft; at *x* = 20 ft it is at maximum displacement
of 3 feet. In between at 18 feet you should see it up at 2 feet. Answer is d).

Return to Question 99.

**Your
answer to Q100: **Sorry, your answer is **not **correct.

Help: *Fundamentals of
Sound*, Secs. 1-D, 1-G.

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 100.

**Hint
for Q100: **This problem uses both graphs. You find how *any* point
moves in time from the second graph and use that information to see how a particular
point on the first graph will move as time goes on.Try dividing any cycle into
four segments.

Return to Question 100.

**Your
answer to Q100: **Congratulations, your answer is **correct!**

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

Return to Question 100

**Correct
answer to Q100: **Each period is 8 seconds. To make things easy divide
a cycle of the motion of the medium point at *x* = 12 into four parts,
1) up from a displacement of -3 ft to zero; 2) from zero up to maximum
height of 3 ft; 3) back down to zero; and 4) back down to -3 ft. Each segment
takes one-fourth of a period or 2 seconds. Thus from 2 seconds to 6 seconds
involves just two segments. The point moves from -3 to zero and then to +3 feet.
Thus the answer is e).

Return to Question 100

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