Respuesta :
Answer:
Tension, T = 2038.09 N
Explanation:
Given that,
Frequency of the lowest note on a grand piano, f = 27.5 Hz
Length of the string, l = 2 m
Mass of the string, m = 440 g = 0.44 kg
Length of the vibrating section of the string is, L = 1.75 m
The frequency of the vibrating string in terms of tension is given by :
[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}[/tex]
[tex]\mu=\dfrac{m}{l}[/tex]
[tex]\mu=\dfrac{0.44}{2}=0.22\ kg/m[/tex]
[tex]T=4L^2f\mu[/tex]
[tex]T=4\times (1.75)^2\times (27.5)^2 \times 0.22[/tex]
T = 2038.09 N
So, the tension in the string is 2038.09 N. Hence, this is the required solution.
The tension is needed to tune this string properly is Tension, T = 2038.09 N
Given information:
Frequency of the lowest note on a grand piano, f = 27.5 Hz
Length of the string, l = 2 m
Mass of the string, m = 440 g = 0.44 kg
The length of the vibrating section of the string is, L = 1.75 m
Calculation of the tension:
[tex]T = 4L^2f\mu\\\\= 4\times (1.75)^2 \times (27.5)^2 \times 0.22[/tex]
T = 2038.09 N
Find out more information about the mass here : brainly.com/question/15959704?referrer=searchResults
