TUTORIAL ANSWERS 130-140

 
 
 


 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?

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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 f0 = 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 f0 = 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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