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Step-by-Step Solution
Step 1: Identify the Physical Quantities Involved
• We have a wire of length l = 1 m, density
\rho_{\text{wire}} = 9 \times 10^{-3}\,\text{kg\,cm}^{-3} .
• The resulting strain in the wire is
4.9 \times 10^{-4} .
• Young’s modulus of the wire,
Y = 9 \times 10^{10}\,\text{N\,m}^{-2} .
• We need to find the lowest frequency (fundamental frequency), denoted by f.
Step 2: Convert Density to SI Units
The given density is in kg/cm³. We must convert it to kg/m³.
1 cm³ = 10^{-6} m³, so
\[
\rho_{\text{wire}}
= 9 \times 10^{-3}\,\text{kg\,cm}^{-3}
\times \frac{1}{\,10^{-6}\,\text{m}^{3}/\text{cm}^{3}}
= 9 \times 10^{-3} \times 10^{6}\,\text{kg\,m}^{-3}
= 9000\,\text{kg\,m}^{-3}.
\]
Step 3: Express the Tension in Terms of Young’s Modulus and Strain
The tension T in the wire (due to stretching) can be found from:
\[
\text{Stress} = Y \times \text{Strain} \quad\implies\quad
\frac{T}{A} = Y \times \left(\text{strain}\right).
\]
Hence,
\[
T = Y \, \times (\text{strain}) \, \times A.
\]
Here A is the cross-sectional area of the wire. Note that when we take
\mu = \rho_{\text{wire}} \times A (mass per unit length), the area A will
cancel appropriately in the formula for frequency.
Step 4: Use the Formula for the Fundamental Frequency of a Stretched String
For a string of length l under tension T, the fundamental frequency is:
\[
f = \frac{1}{2l} \sqrt{\frac{T}{\mu}},
\]
where \mu is the linear mass density = \rho_{\text{wire}} \times A .
Step 5: Substitute T and μ into the Frequency Formula
Substituting
T = Y \times (\text{strain}) \times A and
\mu = \rho_{\text{wire}} \times A ,
\[
f = \frac{1}{2l} \sqrt{\frac{Y \times \text{strain} \times A}{\rho_{\text{wire}} \times A}}
= \frac{1}{2l} \sqrt{\frac{Y \times \text{strain}}{\rho_{\text{wire}}}}.
\]
Step 6: Plug in the Numerical Values
Now, let
l = 1\,\text{m} ,
Y = 9\times 10^{10}\,\text{N\,m}^{-2} ,
strain = 4.9 \times 10^{-4} , and
\rho_{\text{wire}} = 9000\,\text{kg\,m}^{-3} .
So,
\[
f = \frac{1}{2 \times 1}
\sqrt{ \frac{(9 \times 10^{10}) \times (4.9 \times 10^{-4})}{9000} }.
\]
Step 7: Calculate the Frequency
First, multiply out the numerator:
\[
(9 \times 10^{10}) \times (4.9 \times 10^{-4})
= 9 \times 4.9 \times 10^{10 - 4}
= 44.1 \times 10^6
= 4.41 \times 10^7.
\]
Then divide by 9000:
\[
\frac{4.41 \times 10^7}{9000}
= \frac{4.41 \times 10^7}{9 \times 10^3}
= 4.41 \times 10^7 \times 10^{-3} / 9
= \frac{4.41 \times 10^4}{9}.
\]
Approximating,
\[
\frac{4.41 \times 10^4}{9} \approx \frac{4.41}{9} \times 10^4
\approx 0.49 \times 10^4
= 4.9 \times 10^3.
\]
So inside the square root we have approximately 4.9 \times 10^3 ,
and thus
\[
\sqrt{4.9 \times 10^3}
\approx \sqrt{4.9} \times 10^{1.5}
\approx 2.21 \times 10^{1.5}.
\]
In simpler terms, a direct calculation gives roughly:
\[
\sqrt{4.9 \times 10^3} \approx 70.
\]
Therefore,
\[
f
= \frac{1}{2} \times 70 = 35\,\text{Hz}.
\]
Rounding to the nearest integer still gives
f \approx 35\,\text{Hz} .
Final Answer
The lowest frequency of transverse vibrations in the wire is
35 Hz.
