Question

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 594 N to 554.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously

Answer 1

Answer:

3.77 Hz

Explanation:

The beat frequency that is heard can be calculated using the following equation:

$f_{b} = f_{1} - f_{2}$

Where:

f₁ = 110 Hz

$f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}}$

With:

T₁ = 594 N        

T₂ = 554 N

$f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}} = \frac{110 Hz}{\sqrt{\frac{594 N}{554 N}}} = 106.23 Hz$

Hence, the beat frequency is:        

$f_{b} = f_{1} - f_{2} = 110 Hz - 106.23 Hz = 3.77 Hz$      

Therefore, the beat frequency that is heard when the hammer strikes the two strings simultaneously is 3.77 Hz.

I hope it helps you!