Respuesta :
Answer:
B. d(low)=4d(high)
Explanation:
Frequency of a string can be written as;
f = v/2L
Where;
v = sound velocity
L = string length
Frequency can be further expanded to;
f = v/2L = (1/2L)√(T/u) ......1
Where;
m= mass,
u = linear density of string,
T = tension
p = density of string material
A = cross sectional area of string
d = string diameter
u = m/L .......2
m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f ~ 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
The diameter of lowest string is 4 times the diameter of highest string. Hence, option (B) is correct.
Given data:
The number of strings is, 6.
The highest string vibrates with a frequency that is 4 times that of the lowest string.
The expression for the frequency of a string is,
f = v/2L
Here;
v is the sound velocity
L is the string length
And frequency can be further expanded to;
[tex]f=\dfrac{1}{2L}\sqrt{\dfrac{T}{u}}[/tex]
here,
u is the linear density of string, and its value is, u = m/L
m is the mass,
T is the tension in the string.
If p is the density of string material , A is cross sectional area of string and d is string diameter. Then the mass of string is,
m = p AL
[tex]m = \rho \times \dfrac{\pi}{4} \times d^2 \times L[/tex]
Then the frequency is calculated as,
f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))
f = (1/2L)√(T/((p(πd^2)L/4)/L))
f = (1/2L)√(4T/pπd^2)
f = (1/L)(1/d)√(4T/pπ)
Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.
f = 1/d
So, if
4f(low) = f(high)
Then,
d(low) = 4d(high)
Thus, we can conclude that the diameter of lowest string is 4 times the diameter of highest string. Hence, option (B) is correct.
Learn more about the frequency here:
https://brainly.com/question/8304265
