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1.) In the figure, suppose that the separation between speakers A and B is 4.30 m and the speakers are vibrating in phase. They are playing identical 130-Hz tones, and the speed of sound is 343 m/s. What is the largest possible distance between speaker B and the observer at C, such that he observes destructive interference?
m

2.) Two speakers, one directly behind the other, are each generating a 225-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is 339 m/s.
m

3.) Two pure tones are sounded together. The drawing shows the pressure variations of the two sound waves, measured with respect to atmospheric pressure, where t1 = 0.0202 s and t2 = 0.0250 s. What is the beat frequency?
Hz

4..) Two ultrasonic sound waves combine and form a beat frequency that is in the range of human hearing for a healthy young person. The frequency of one of the ultrasonic waves is 89 kHz. (The range of hearing for a healthy young person is 20 Hz to 20 kHz.) (a) What is the smallest possible value for the frequency of the other ultrasonic wave?
kHz

(b) What is the largest possible value for the frequency of the other ultrasonic wave?

(a) Determine the speed of the traveling wave.
m/s

(b) Determine the wavelength of the traveling wave.
m

(c) Determine the frequency of the traveling waves that make up the standing wave.
Hz

6.) A 40-cm length of wire has a mass of 6.0 g. It is stretched between two fixed supports and is under a tension of 165 N. What is the fundamental frequency of this wire?
Hz

8.) The string on a string bass vibrates at a fundamental frequency of 77.0 Hz. If the string's tension were increased by a factor of 4, what would be the new fundamental frequency?
Hz

Explanation / Answer

8) The frequency is proportional to the square root of the tension, so increasing the tension by a factor of four would double the frequency.i.e is 164 HZ
7) If T is the tension, M is the linear density, and v is the speed, then

v = ?(T/M).

So the speed of the wave is ?(3.2/(5.95x10^-7) = 2319.08 m/s.

Now with any wave, v = f? speed = frequency times wavelength. So ? = v/f = 16.68 is the wavelength of the wave.

Now, we are at the third harmonic frequency. The string is fixed at both ends, so therefore it must be that the string is an exact multiple of half a wavelength long.

The first harmonic is when the string is exactly half a wavelength long. The second one is when the string is exactly 1 full wavelength long. The third one is when the string is exactly 1.5 wavelengths long,

So the string must be 1.5*16.68 = 25.02 m long.
6)

     V = ? T / (m/L) = ? 165 / (0.006 / 0.4 ) = 104.88 m/s

f = c /? = 104.88 / 0.96 = 109.25 Hz
4) A. Smallest is 89 kHz - 20 kHz = 69 kHz
B. Largest is 89 kHz + 20 kHz = 109 kHz

3) f1 = 1/t1 = 1/0.0202 = 49.504 Hz
f2 = 1/t2 = 1/0.0250 = 40 Hz

No. of beats =9.504 appx

2) lambda=v/f or 339/225 = wavelength (1.506m) the speakers would have to be setup at 1/2 wavelength or 0.75m for destructive interference.

1) Destructive interference occurs when two equal phase sources are observed with a path length difference of a half wavelength. The wavelength of the sound is:

? = v / f = 343 / 130 = 2.638 m

Let the distance to speaker B be d, and the separation between the two speakers be s = 4.3 m. Calculate the length to speaker A:

CA = sqrt( d^2 + s^2)

Set CA - d = n*?/2, n = 1, 3, 5, 7, ...

sqrt(d^2 + s^2 ) - d = ?/2

sqrt( d^2 + s^2 ) = ?/2 + d

d^2 + s^2 = ?^2/4 + ?*d + d^2

( s^2 - ?^2/4 ) / ? = d

d = (4.3^2 - 2.63^2/4 ) / 2.63 = 6.36 m
5) a) v = sqrt(230/5.8x10^-3) = 199.1
  b) Two loops of any standing wave comprise one
wavelength. Since the string is 2.5 m long and consists
of three loops , the wavelength is 2/3 x 2.5 =1.66
c) f = 199.1/1.66 =119.93

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