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Two identical wires A and B are subject to tension. The tension in wire A is 3 times larger than that in wire B. Find the ratio of the frequencies of the first harmonic in these two wires, fA1 / fB1.

Respuesta :

lublana

Answer:

1.732

Explanation:

Let

Tension in wire B=T

Tension in wire A=3 T

We have to find the ratio of the frequencies of the first harmonic in these two wires.

When two wires are identical then the length of both wires are same.

Suppose, the length of each wire=l

Frequency=[tex]\frac{1}{2l}\sqrt{\frac{T}{\mu}}[/tex]

Where [tex]\mu=[/tex]Mass per unit length

Mass per unit length of both wires are same because the two wires are identical.

[tex]\mu_A=\mu_B[/tex]

[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_B}}}[/tex]

[tex]\frac{f_A}{f_B}=\frac{\frac{1}{2l}\sqrt{\frac{T_A}{\mu_A}}}{\frac{1}{2l}\sqrt{\frac{T_B}{\mu_A}}}=\sqrt{\frac{T_A}{T_B}}=\sqrt{\frac{3T}{T}}[/tex]

[tex]\frac{f_A}{f_B}=\sqrt 3=1.732[/tex]

onyebuchinnaji

The ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].

The given parameters;

  • tension in wire A = T
  • tension in wire B = 3T

The frequency of first harmonic of each wire is calculated as follows;

[tex]F_ A = \frac{1}{2l} \sqrt{\frac{3T}{\mu} } \\\\F_B = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

  • l is the length of the wires
  • T is the tension on the wire
  • μ is the mass per unit length

The ratio of the two frequencies is calculated as follows;

[tex]\frac{F_A}{F_B} = \frac{\frac{1}{2l} \sqrt{\frac{3T}{\mu} } }{\frac{1}{2l} \sqrt{\frac{T}{\mu} } } \\\\\frac{F_A}{F_B} = \sqrt{\frac{3T}{T} } \\\\\frac{F_A}{F_B} = \sqrt{3}[/tex]

Thus, the ratio of the frequencies of the first harmonic in these two wires is [tex]\sqrt{3}[/tex].

Learn more here:https://brainly.com/question/14149129

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