1. Two in-phase loudspeakers are some distance apart. They emit sound with a fre
ID: 1687657 • Letter: 1
Question
1. Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 1536 Hz. You move between the speakers, along the line joining the them, at a constant speed of 2.8 m/s. What beat frequency do you observe? The speed of sound in the room is 330 m/s. 2. In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?1. Two in-phase loudspeakers are some distance apart. They emit sound with a frequency of 1536 Hz. You move between the speakers, along the line joining the them, at a constant speed of 2.8 m/s. What beat frequency do you observe? The speed of sound in the room is 330 m/s. 2. In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?
Explanation / Answer
1. For the first question it is asking for the beat frequency. V_s =330m/s and V_L= 2.8 m/s The equation you want to use is f_(L)=(V+V_L)/V_S
Since you want it between them you have f_L= ((V_L/V_S)+1)*f
and f_L=((V_L/V_S)-1)f
Then you use the f_beat=f_a-f_b.
With numbers:
(2.8)/330 = 8.48*10^-3
(1+ans)*1536=1549.03
(2.8/330) = 8.48*10^-3
(1-ans)*1536 = 1523
1549-1523=26
Hope this helps!
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