TUTORIAL ANSWERS 300-330

 

 
 

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/2L. 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 2f0 , 4f0, 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.

 

 


 

 


ab_webmaster@abacon.com
©2002 William J. Mullin
Legal Notice