Answer:
A. 4.0%
Explanation:
The value of the frequency, f, in a string with a tension, T, is:
[tex]f=\sqrt{\frac{T}{4mL} }[/tex]
For the well calibrated instrument:
[tex]f_{1}=\sqrt{\frac{T_{1}}{4mL} }[/tex]
For the instrument to be calibrated
[tex]f_{2}=\sqrt{\frac{T_{2}}{4mL} }[/tex]
We divide these equations in order to find the rapport between the Tensions:
[tex]T_{2}/T_{1}=(f_{2}/f_{1})^{2}=(448.4/440)^2=1.04[/tex]
if the rapport between T2 y T1 is 1.04, it means that the tension in the string should be changed in 4%.
