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 610 N to 542.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?

To calculate the beat frequency, we first need to determine the frequency of the slipped string. We can use Hooke's Law to find the relationship between tension, frequency, and length. Hooke's Law states that the frequency of a string is inversely proportional to the square root of its tension.

Let's start by calculating the frequency of the string before it slipped.

Given:
Tension before slipping (T1) = 610 N
Frequency before slipping (f1) = 110 Hz

We can assume that the tension is directly proportional to the square of the frequency, and therefore:

(T1 / T2) = (f2 / f1)^2

Rearranging the equation, we can solve for the frequency of the slipped string (f2):

f2 = f1 * sqrt(T1 / T2)

Plugging in the values:

f2 = 110 Hz * sqrt(610 N / 542.00 N)

f2 ≈ 110 Hz * sqrt(1.1279)

f2 ≈ 110 Hz * 1.0625

f2 ≈ 116.88 Hz

So, the frequency of the slipped string is approximately 116.88 Hz.

Now, we can calculate the beat frequency, which is the absolute value of the difference between the frequencies of the two strings:

Beat frequency = |f1 - f2|

Beat frequency = |110 Hz - 116.88 Hz|

Beat frequency ≈ 6.88 Hz

Therefore, when the hammer strikes the two strings simultaneously, the beat frequency heard is approximately 6.88 Hz.