A uniform string of length L and mass M is fi | vedclass

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Question

(A) The fundamental frequency of a string fixed at both ends is given by $v = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n$ is the harmonic number, $L

Vedclass · Accepted Answer

$v=\frac{n}{2} \sqrt{\frac{T}{M L}}$

Vedclass · Answer

(A) The fundamental frequency of a string fixed at both ends is given by $v = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$, where $n$ is the harmonic number, $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.Linear mass density $\mu = \frac{M}{L}$.Substituting $\mu$ into the formula:$v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2L} \sqrt{\frac{TL}{M}}$.Wait, let us re-evaluate: $v = \frac{n}{2L} \sqrt{\frac{T}{M/L}} = \frac{n}{2} \sqrt{\frac{T}{M/L \cdot L^2}} = \frac{n}{2} \sqrt{\frac{T}{ML}}$.Thus, the correct formula is $v = \frac{n}{2} \sqrt{\frac{T}{ML}}$.

$A$ uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1, 2, 3, \ldots$):

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