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den301095 [7]
3 years ago
6

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

e 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?
Physics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

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

<u>f = 5.65 Hz</u>

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