Musicians can use beats to tune their instruments. One flute is properly tuned and plays the musical note A at exactly 440 Hz. A second player sounds the same note and hears that her instrument is slightly "flat" (that is, at too low a frequency). Playing at the same time as the first flute, she hears two loud-soft-loud beats per second. What is the frequency of her instrument?

The frequency of the second flute is 437Hz

Explanation:

Sound is a wave that travels in a medium as longitudinal waves. Just like any other coherent beams of waves, sound waves also can interfere with each other given that they exist in the same medium. A beat is the result of the interference of sound, this is a combination of one high-intensity sound and the following low-intensity sound. Beats can be used to determine the difference in the frequencies of the sound waves since they emerge due to the difference of the frequencies of interfering waves.

Given:

Frequency of the properly tuned F1F1) = 440 HzBeat characteristics = 3 loud-soft-loud beats per second

The given beat characteristic is just the frequency of the beat produced by the interfering sound waves. Beats are described as the difference in the frequency of the waves. We can describe the beat , b

The given beat characteristics

b = F1 - F2

where F2 is the unknown frequency of the second flute.

The equation already considers that the second player has a flat flute with a lower frequency

Solving for the second flute:

F2 = F1 - b

F2 = 440 - 3

F2 = 437Hz

The frequency of her instrument is 438 Hz

Explanation:

Considering the beat frequency of 2 waves is

fbeat = f1 - f2

Rewriting this equation we can get (considering lower frequency - use the minus  sign)

f2 = f1 - fbeat = 440 - 2 = 438 Hz

this is a good question! i'll do some research and come back when i have the answer!

d. platinum would have the greatest temperature increase because it has the lowest specific heat.