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A wire vibrates at a fundamental frequency of $500 \mathrm{~Hz}$. A second identical wire produces 5 beats per second with it when the tension in the first wire is slightly decreased. The ratio of the tension in the second wire to the tension in the first wire is approximately equal to
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The correct answer is:
1.02
(d) Given that, fundamental frequency of first wire, $f_1=500 \mathrm{~Hz}$
Let frequency of 2 nd wire be $f_2$.
Beat frequency, $f_b=5$
$$
\begin{aligned}
& f_b=f_2-f_1 \\
& f_2=f_b+f_1=500+5=505 \mathrm{~Hz}
\end{aligned}
$$
Using, expression for fundamental frequency of wire,
$$
\begin{aligned}
f & =\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow f \propto \sqrt{T} \\
\Rightarrow \quad \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{505}{500}=\sqrt{\frac{T_2}{T_1}} \\
\Rightarrow \quad \frac{T_2}{T_1} & =\left(\frac{505}{500}\right)^2=1.02
\end{aligned}
$$
Let frequency of 2 nd wire be $f_2$.
Beat frequency, $f_b=5$
$$
\begin{aligned}
& f_b=f_2-f_1 \\
& f_2=f_b+f_1=500+5=505 \mathrm{~Hz}
\end{aligned}
$$
Using, expression for fundamental frequency of wire,
$$
\begin{aligned}
f & =\frac{1}{2 L} \sqrt{\frac{T}{\mu}} \Rightarrow f \propto \sqrt{T} \\
\Rightarrow \quad \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \Rightarrow \frac{505}{500}=\sqrt{\frac{T_2}{T_1}} \\
\Rightarrow \quad \frac{T_2}{T_1} & =\left(\frac{505}{500}\right)^2=1.02
\end{aligned}
$$
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