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?

f = 5.65 Hz

Explanation:

The fundamental frequency of a string is given by following formula:

f = v/2L

where,

f = fundamental frequency

v = speed of wave = √(TL/m)

L = Length of String

m = Mass of String

T = Tension in String

Therefore,

f = √(TL/m)/2L

2f = √(T/Lm)

For initial condition:

T₁ = 600 N

f₁ = 110 Hz

2(110 Hz) = √(600 N/Lm)

√(600 N)/220 Hz =  √Lm

Lm = 0.01239 N/s²

Now, for changed tension:

2f₂ = √(540 N/0.01239 N/s²)

f₂ = 208.7 Hz/2

f₂ = 104.35 Hz

So, the beat frequency will be:

f = f₁ - f₂

f = 110 Hz - 104.35 Hz

f = 5.65 Hz

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