Question 447: Waves

Question

A uniform metal wire has length $L$, mass $M$ and density $\rho$. It is under tension $T$ and $v$ is the speed of transverse wave along the wire. The area of cross-section of the wire is

Accepted Answer

To determine the area of the cross-section of a uniform metal wire given its length $L$, mass $M$, density $ho$, tension $T$, and the speed $v$ of a transverse wave along it, we use the following relationship:
The speed of a transverse wave along a wire is described by:
$ v = \sqrt{\frac{T}{\mu}} $
where $\mu$ represents the mass per unit length of the wire. The mass per unit length $\mu$ can be expressed in terms of the wire's density and cross-sectional area:
$ \mu = 	ext{volume per unit length} 	imes 	ext{density} = 	ext{area} 	imes 	ext{density} = A 	imes ho $
Substituting back into the wave speed equation gives:
$ v = \sqrt{\frac{T}{A ho}} $
Solving for the area $A$, we obtain:
$ A = \frac{T}{v^2 ho} $
Thus, the area of the cross-section of the wire is $\frac{T}{v^2 ho}$.

Answer

$\frac{T}{v^2 \rho}$

Answer

$\frac{v^2 \rho}{T}$

Answer

$T^2 \rho V$

Answer

$T v^2 \rho$

Waves

A uniform metal wire has length L, mass M and density \rho. It is under tension T and v is the speed of transverse wave along the wire. The area of cross-section of the wire is

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