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(D) The fundamental frequency of a stretched wire is given by $\mu = \frac{1}{2L} \sqrt{\frac{F}{\lambda}}$,where $\lambda$ is the mass per unit lengt

Vedclass · Accepted Answer

$\frac{1}{2}\sqrt{\frac{F}{ML}}$

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(D) The fundamental frequency of a stretched wire is given by $\mu = \frac{1}{2L} \sqrt{\frac{F}{\lambda}}$,where $\lambda$ is the mass per unit length.Given that the total mass is $M$ and length is $L$,the mass per unit length $\lambda = \frac{M}{L}$.Substituting this into the frequency formula:$\mu = \frac{1}{2L} \sqrt{\frac{F}{M/L}}$$\mu = \frac{1}{2L} \sqrt{\frac{FL}{M}}$$\mu = \frac{1}{2} \sqrt{\frac{FL}{M \cdot L^2}}$$\mu = \frac{1}{2} \sqrt{\frac{F}{ML}}$Resonance occurs when the impressed frequency matches the fundamental frequency of the wire,leading to maximum vibrational energy.

$L$ length and $M$ mass uniform wire is stretched between two fixed points with tension $F$. $A$ sound of frequency $\mu$ is impressed on it. The maximum vibrational energy exists in the wire when $\mu$ =

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