A uniform wire of length $L$ and mass $M$ is stretched between two fixed points, keeping tension $F$. A sound of frequency $m$ is impressed on it. Then the maximum vibrational energy is existing in the wire when $\mu $ =
Diffcult
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The frequency of the wire depends on the length of the string $(L),$ tension $(F)$ and the mass per unit length
of the wire $(\mathrm{m}=\mathrm{M} / \mathrm{L})$
Let $\left.A \propto L^{a} F^{b} m^{c} \quad \text { ……….. } \quad \text { (Equation } 1\right)$
Writing the dimensions of each in the above equation, we get
$\left[\mathrm{M}^{0} \mathrm{L}^{0} \mathrm{T}^{-1}\right]=\mathrm{L}^{\mathrm{a}}\left[\mathrm{M}^{1} \mathrm{L}^{1} \mathrm{T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{M}^{0} \mathrm{L}^{-1}\right]^{\mathrm{C}}$
$=M^{b+c} L^{a+b-c} T^{-2 b}$
Here, $b+c=0$
$a+b-c=0 \text { and }$
$-2 b=-1$
On solving, we get
$a=-1, b=1 / 2$ and $c=-1 / 2$
Substituting these values in equation $(1),$ we get
$A=k L^{-1} F^{1 / 2} m^{-1 / 2}$
$=k \frac{1}{L} \sqrt{\frac{F}{m}}$
or $A=k \frac{1}{L} \sqrt{\frac{F}{(M / L)}}$
or $A=k \sqrt{\frac{F}{M L}}$
Experimentally $\mathrm{k}=1 / 2$ $\therefore A=\frac{1}{2} \sqrt{\frac{F}{M L}}$
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