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lidiya [134]
3 years ago
11

Inside a spinnet, the hammer is sometimes set to strike multiple strings to increase the instrument's volume. If the 523 Hz note

has two strings, and one slips from its normal tension of 620 N to 610 N, what beat frequency in Hz is heard when the note is played?
Physics
1 answer:
rodikova [14]3 years ago
3 0

Answer:

4.23 beats / seconds

Explanation:

Frequency heard = f = f₁ - f₂

f₂ = f₁  √ ( F₂ / F₁)

f = f₁ - f₁ √ ( F₂ / F₁) = f₁ ( 1 - √ ( F₂ / F₁) = 523 Hz ( 1 - 0.9919 ) = 4.23 beats / seconds

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