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(end of section 18.1)

QUICK QUIZ 18.1.

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(end of section 18.1)

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  1. QUICK QUIZ 18.1 In Example 18.1, we discovered that a listener at point P would hear a minimum in the sound when the oscillator driving both speakers was at a frequency of 1.3 kHz. At point O, on the other hand, the listener would hear a maximum in the sound. The intensity of the sound heard at point P would be a) less than but close to the intensity of the sound heard at point O, b) half the intensity of the sound heard at point O, c) very faint but not zero, or d) zero. (end of section 18.1)

  2. QUICK QUIZ 18.1 ANSWER (c). The two sound waves from each speaker would destructively interfere at point P. Recall, however, that the amplitude would be very slightly less from speaker 2 than from speaker 1 at this point because of the longer distance to speaker 2, and the 1/r2 dependence of the intensity. The two slightly different amplitudes would not completely cancel when summed, as illustrated below.

  3. QUICK QUIZ 18.2 (end of section 18.3) • The quantity that is the same for all six strings on a guitar is a) the fundamental frequency, b) the fundamental wavelength, c) the speed of a wave on the strings, d) all of the above, or e) none of the above.

  4. QUICK QUIZ 18.2 ANSWER (b). The length between the two fixed ends (the bridge at the top of the neck and at the other end of the guitar) is the same for each guitar string. Since the fundamental wavelength for a string fixed at both ends is just double this length, the fundamental wavelength is the same for all six strings.

  5. QUICK QUIZ 18.3 (end of section 18.3) In Example 18.5, by immersing the hanging sphere in water, the buoyant force changes the tension in the string so that the string vibrates in its fifth harmonic. Suppose you would like to make the string vibrate in its first harmonic. Instead of immersing the sphere in water, you should a) double the mass of the hanging sphere, b) triple the mass of the hanging sphere, c) quadruple the mass of the hanging sphere, or d) increase the mass of the hanging sphere by a factor of eight.

  6. QUICK QUIZ 18.3 ANSWER (c). As described in Example 18.5, the ratio of successive harmonic numbers is determined by, n2/n1 = (T1/T2) or (T1/T2) = (n2/n1)2. Therefore, the tension should be quadrupled if the harmonic number goes from 2 to 1.

  7. QUICK QUIZ 18.4 (end of section 18.5) • For a musical instrument such as the flute, as the temperature of the air increases, both the velocity of sound in air increases and the flute slightly increases in length. The fact that the flute becomes more sharp with increasing temperature implies that a) the sharpness has nothing to do with the length of the flute, b) the percentage increase in length is greater than the percentage increase of the velocity of sound in air, c) the percentage increase in length is about the same as the percentage increase of the velocity of sound in air, or d) the percentage increase in length is less than the percentage increase of the velocity of sound in air.

  8. QUICK QUIZ 18.4 ANSWER (d). Equation 18.11, fn = n(v/2L), indicates that an increasing length will contribute to a lower frequency while an increase in the velocity will contribute to a higher frequency. The fact that the frequency is higher (the instrument is sharp) indicates that the percentage change in the length is less than the percentage change in the velocity.

  9. QUICK QUIZ 18.5 You simultaneously hit the lowest G key on the piano and the G key that is one octave higher (an octave corresponds to a doubling of the frequency). The frequency of the beats that you hear will be a) the same as the fundamental frequency of the lowest G key, b) the same as the fundamental frequency of the higher G key, c) twice the fundamental frequency of the lowest G key, d) twice the fundamental frequency of the higher G key, or e) impossible to determine without knowing what the fundamental frequency of the lowest G key is. (end of section 18.7)

  10. QUICK QUIZ 18.5 ANSWER (a). We can use Equation 18.15, fb = |f1-f2|, with f2 = 2f1 which yields, fb= |f1-2f1| = f1. Since the beat frequency is the same as the frequency of the lowest note, the sound of the beats will complement the lowest note. This explains why notes of successive octaves sound good together.

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