Question

Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency, 440 Hz. The other is producing a frequency of 448.4 Hz. By what percentage should the out-of-tune musician change the tension in his string to bring his instrument into tune at 440 Hz?

A. 4.0%
B. 0.5%
C. 2.0%
D. 1.0%
E. 0.6%

Answer 1

The musician should decrease the tension in their violin string by approximately 2.0% to change the frequency from 448.4 Hz to the desired 440 Hz.

Step-by-step explanation:

The out-of-tune musician is producing a frequency of 448.4 Hz, while the desired frequency is 440 Hz. The frequency of a string is related to the tension by the formula f = (1/2L) ∙ √(T/μ), where f is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density. To tune the violin, the musician should decrease the frequency to match 440 Hz, which is a decrease in frequency of 448.4 Hz - 440 Hz = 8.4 Hz.

To find the percentage change, we divide the change in frequency by the initial frequency: (8.4 Hz / 448.4 Hz) × 100% = 1.87%, which is approximately 2.0%. Therefore, the musician should adjust the tension by this percentage to bring the violin into tune.

Answer 2

A. 4.0%

Step-by-step explanation:

The value of the frequency, f, in a string with a tension, T, is:

`f=\sqrt{(T)/(4mL) }`

For the well calibrated instrument:

`f_(1)=\sqrt{(T_(1))/(4mL) }`

For the instrument to be calibrated

`f_(2)=\sqrt{(T_(2))/(4mL) }`

We divide these equations in order to find the rapport between the Tensions:

`T_(2)/T_(1)=(f_(2)/f_(1))^(2)=(448.4/440)^2=1.04`

if the rapport between T2 y T1 is 1.04, it means that the tension in the string should be changed in 4%.