Question

String A has a length I, radius of cross – section r, density of material $\rho$ and is under tension t. string B has all these quantities double those of string A .if $\upsilon_{A}$ and $\upsilon_{B}$ are the corresponding fundamental frequencies of the vibrating string, then

• A. $\upsilon_{A} = 2\upsilon_{B}$
• B. $\upsilon_{A} = 4\upsilon_{B}$
• C. $\upsilon_{B} = 4\upsilon_{A}$
• D. $\upsilon_{A} = \upsilon_{B}$

Answer 1

Answer: (B)

$\upsilon_{A} = 4\upsilon_{B}$

Answer 2

Fundamental frequency of a string is

$\upsilon = \frac{1}{2L}\sqrt{\frac{T}{\pi r^{2}\rho}} = \frac{1}{2Lr}\sqrt{\frac{T}{\pi}}$

Where the symbols have their usual meanings.

$\therefore\frac{\upsilon_{A}}{\upsilon_{B}} = \left( \frac{L_{B}}{L_{A}} \right)\left( \frac{r_{B}}{r_{A}} \right)\left( \frac{T_{A}}{T_{B}}\frac{\rho_{B}}{\rho_{A}} \right)^{1/2}$

Substituting the given values, we get

$\frac{\upsilon_{A}}{\upsilon_{B}} = \left( \frac{2L}{L} \right)\left( \frac{2r}{r} \right)\left( \frac{T}{2T}\frac{2\rho}{\rho} \right)^{1/2} = 4$

$\upsilon_{A} = 4\upsilon_{B}$