Question

Two identical strings on different cellos are tuned to the 440-hertz A note. The peg holding one of the strings slips, so its tension is decreased by 1.7%. What is the beat frequency heard when the strings are then played together?

Heads up I recently posted this same question and the expert answere given was 434.28. This answere is inncorect.

Answer 1

Beat is just the difference between the two frequencies , (f_new – f_old).

Given that

(Τ_new – Τ_old) / (Τ_old) = 0.017

From the relation Τ = μv²:

(μv_new² – μv_old²) / μv_old² = 0.017

From the relation v = λf:

(μ(λf_new)2 – μ(λf_old)2) / μ(λf_old)2 = 0.017

(f_new² – f_old²) / f_old² = 0.017

f_new² = 0.017f_old² + f_old²

f_{new =} √ (0.017f_old² + f_old² )

f_{new =} √ (0.017 (440)² + (440)² )

f_{new =} √ (0.017 (440)² + (440)² )

f_{new =} 443.72 Hz

Therefore, the beat frequency

= 443.72 Hz - 440 Hz = 3.72 Hz

In two significant figures, beat frequency is 3.7 Hz