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Nat2105 [25]
4 years ago
12

Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s.A violinist is tuning her instrument to concert A

(440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat of frequency 6.00 Hz, which increases to 7.00 Hz when she tightens her violin string slightly.
(a) What was the frequency of the note played by her violin when she heard the 3-Hz beats?
(b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?
Physics
1 answer:
Molodets [167]4 years ago
7 0

Answer:

A) 443 Hz

B) She has to loosen the string

Explanation:

A) Given;

Beat frequency;f_beat = 3 Hz

Frequency of electronically generated tone; f_e = 440 Hz

We know that formula for beat frequency is given by;

f_beat = |f1 - f2|

Now, applying it to this question, we have;

f_beat = f_v - f_e

Where f_v is frequency of the note played by the violinist

Thus, plugging in the relevant values;

3 = f_v - 440

f_v = 3 + 440

f_v = 443 Hz

B) In the concept of wave travelling in a string, the frequency is directly proportional to the square root of the force acting on the string.

Now, for the violinist to get her violin perfectly tuned to concert A from what it was when she heard the 3-Hz beats, the beat frequency will have to be zero. Which means it has to decrease by 3 Hz. For it to decrease, it means that the force applied has to decrease as we have seen that frequency is directly proportional to the square root of the force acting on the string.

Thus, she would have to loosen the string.

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