Consider the two identical piano strings, each tuned exactly to the $420\ \mathrm{Hz}$. The tension in any…
Consider the two identical piano strings, each tuned exactly to the $420\ \mathrm{Hz}$. The tension in any one of the strings is increased by $2.0\%$. If they are now struck, what is the beat frequency between the fundamentals of the two strings? (Take, length of the strings = $65\ \mathrm{cm}$)
Solution
Sol. If $\nu_1, v_1, T_1$ and $\nu_2, v_2$ and $T_2$ are the frequencies, velocities and tensions in the first and second strings respectively, then
$\dfrac{v_2}{v_1}=\dfrac{v_2/2L}{v_1/2L}=\dfrac{v_2}{v_1}$
$\Rightarrow\;\dfrac{\nu_2}{\nu_1}=\dfrac{\frac{1}{2L}\sqrt{T_2/\rho}}{\frac{1}{2L}\sqrt{T_1/\rho}}=\sqrt{\dfrac{T_2}{T_1}}$
Since, it is given that the tension in one string is $2\%$ larger than the other.
$\therefore\;T_2=T_1+\dfrac{2T_1}{100}=1.02\;T_1$
$\therefore\;\dfrac{\nu_2}{\nu_1}=\sqrt{\dfrac{1.02\;T_1}{T_1}}=1.01$
Now, the frequency of the tightened string,
$\nu_2=\nu_1(1.01)=1.01\times 420$
$=424.2\;\mathrm{Hz}$
$\therefore$ Beat frequency,
$\nu_{beat}=\nu_2-\nu_1=424.2-420$
$=4.2\;\mathrm{Hz}$
Answer: $4.2\ \mathrm{Hz}$