JEE Advance - Physics (2023 - Paper 2 Online - No. 10)
A string of length $1 \mathrm{~m}$ and mass $2 \times 10^{-5} \mathrm{~kg}$ is under tension $T$. When the string vibrates, two successive harmonics are found to occur at frequencies $750 \mathrm{~Hz}$ and $1000 \mathrm{~Hz}$. The value of tension $T$ is ________ Newton.
Answer
5
Explanation
$I=1 \mathrm{~m}, m=2 \times 10^{-5} \mathrm{~kg},$
T : Tension in the string.
$\because$ Successive frequencies are being given
$\therefore$ It is the case of both ends fixed.
Now,
$$ \begin{aligned} & f_{n+1}-f_n=1000-750 \\\\ \Rightarrow & \frac{(n+1)}{2 l} \sqrt{\frac{T}{\mu}}-\frac{n}{2 l} \sqrt{\frac{T}{\mu}}=250 \\\\ \Rightarrow & \frac{1}{2 l} \sqrt{\frac{T}{\mu}}=250 \end{aligned} $$
$\begin{aligned} & \Rightarrow \sqrt{\frac{T}{2 \times 10^{-5}}}=250 \times 2 \times 1 \\\\ & \Rightarrow \frac{T}{2 \times 10^{-5}}=25 \times 10^{-4} \\\\ & \Rightarrow T=50 \times 10^{-1} \\\\ & T=5 \mathrm{~N}\end{aligned}$
T : Tension in the string.
$\because$ Successive frequencies are being given
$\therefore$ It is the case of both ends fixed.
Now,
$$ \begin{aligned} & f_{n+1}-f_n=1000-750 \\\\ \Rightarrow & \frac{(n+1)}{2 l} \sqrt{\frac{T}{\mu}}-\frac{n}{2 l} \sqrt{\frac{T}{\mu}}=250 \\\\ \Rightarrow & \frac{1}{2 l} \sqrt{\frac{T}{\mu}}=250 \end{aligned} $$
$\begin{aligned} & \Rightarrow \sqrt{\frac{T}{2 \times 10^{-5}}}=250 \times 2 \times 1 \\\\ & \Rightarrow \frac{T}{2 \times 10^{-5}}=25 \times 10^{-4} \\\\ & \Rightarrow T=50 \times 10^{-1} \\\\ & T=5 \mathrm{~N}\end{aligned}$
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