The length,tension,diameter,and density of a | vedclass

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Question

(D) The velocity of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is linear mass density

Vedclass · Accepted Answer

Both $(A)$ and $(C)$.

Vedclass · Answer

(D) The velocity of a transverse wave in a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is linear mass density.Since $\mu = \rho \cdot A_{cross} = \rho \cdot \pi r^2$,we have $\mu \propto \rho \cdot d^2$ (where $d$ is diameter).Given for wire $B$: $L_B = 2L_A$,$T_B = 2T_A$,$d_B = 2d_A$,and $\rho_B = 2\rho_A$.Velocity ratio: $\frac{v_B}{v_A} = \sqrt{\frac{T_B}{T_A} \cdot \frac{\mu_A}{\mu_B}} = \sqrt{\frac{2T_A}{T_A} \cdot \frac{\rho_A d_A^2}{\rho_B d_B^2}} = \sqrt{2 \cdot \frac{1}{2} \cdot \frac{1}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Thus,$v_B = \frac{1}{2} v_A$. So,option $(C)$ is correct.Fundamental frequency $n = \frac{v}{2L}$.$\frac{n_B}{n_A} = \frac{v_B}{v_A} \cdot \frac{L_A}{L_B} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.So,$n_A = 4n_B$. The third overtone of $B$ is $4n_B$,which equals $n_A$. Thus,option $(A)$ is also correct.Therefore,both $(A)$ and $(C)$ are correct.

The length,tension,diameter,and density of a wire $B$ are double those of the corresponding quantities for another stretched wire $A$. Then:

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