Quiz Solutions

Quiz 1 (090607) [10 points; each numbered question is 5 points]

1. The speed of sound in helium at 20°C is about 1007 m/s. An organ pipe that normally [i.e., in air at 20°C] sounds at middle C (approximately 261 Hz) will sound at what frequency when blown with helium?

v = fl, (always, for sound!). Since f is thus proportional to v, we can write

v_He/v_air = f_He/f_air => f_He = (v_He/v_air)f_air = (1007 m/s)/344m/s) (261 Hz)

= 764 Hz

2. What is the wavelength of the above sound when the pipe is

a) blown with air?

v = fl, so l =v/f = (344m/s)/261 Hz = 1.32m

b) blown with helium?

v = fl, so l =v/f =(1007m/s)/764 Hz = 1.32m

Note that the wavelength of the sound wave remained the same. The frequency is the parameter that's affected by the change in velocity!

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Quiz 2 (0918 07)

1. An aerial bomb explodes high in the air above Florida Field during the Gator Growl fireworks display. The sound intensity level of the 'bang' at a point on the ground beyond the stadium, 400m from the explosion is 80 dB.

What is the probable intensity level at a point 800m (about half a mile) away, assuming no blockages or reflections?

Apply the inverse square law: I₁/I₂ = (r₂/r₁)² . Thus

I₂ = I₁(r₁/r₂)² = I₁(400m/800m)² = I₁(0.25). In other words, I₁ is 4 times I₂ .

We don't know either intensity but from the discussion of the inverse square law and Table 5.1, we know that every doubling of intensity is equivalent to a 3dB increase in the sound intensity level (SIL). Conversely, there's a 3dB decrease in the SIL for every halving of intensity. So quadrupling => 2 x 3 dB = 6 dB, and

SIL at 800m is 80 dB - 6 dB = 74 dB

2. As you are tuning your favorite musical instrument to "A440" with the aid of an accurate tuning fork, you hear 4 beats per second.

a) What is the frequency of the instrument's note?

f_B = f₂ -f₁ . This is due to wave interference and the beat frequency is just [(higher frequency) - (lower frequency)]. But unless you know which one (tuning fork or instrument) is higher, there are two possible answers for the instrument's note or frequency:

f = 440 Hz - 4 Hz = 436 Hz or f = 440 Hz + 4 Hz = 444 Hz.

b) How do you determine whether the instrument is sharp (higher f) or flat (lower f)?

Change the instrument's frequency in a known direction. If you increase its frequency and the beats decrease, it was flat; If the beats increase, it was sharp. Similarly, if you decrease the instrument's frequency and the beats decrease, it was sharp; if they increase, it was flat.

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