Question

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?

Answer 1

When the note is played on the spinnet, there will be no beat frequency heard.

Step-by-step explanation:

When two strings are struck simultaneously in a spinnet, beats can be heard. Beats are the result of the interference between two waves with slightly different frequencies. The beat frequency is equal to the difference in frequencies between the two strings.

In this case, one string is tuned to 523 Hz and the other string slips from its normal tension of 620 N to 610 N. To find the beat frequency, we need to calculate the difference in frequencies.

f = f1 - f2 = 523 Hz - 523 Hz = 0 Hz

Therefore, when the note is played, no beat frequency will be heard.

Answer 2

4.23 beats / seconds

Step-by-step 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