Physics

Problem

Two wires A and B of the same material are stretched between the same pair of rigid supports, but they differ in length and linear mass density. For wire A,

L_A = 80\,\text{cm}

and

\mu_A = 0.012\,\text{kg}\,\text{m}^{-1}

. For wire B,

L_B = 50\,\text{cm}

and

\mu_B = 0.020\,\text{kg}\,\text{m}^{-1}

. Each wire is kept under a (possibly different) tension and both are observed to have the same fundamental frequency. a. (i) Derive an expression connecting the tensions

T_A

and

T_B

in terms of the given lengths and mass densities, based on the condition of equal fundamental frequencies. b. (ii) Using the given numerical values, calculate the ratio

T_A : T_B

. c. (iii) If the tension in wire A is

120\,\text{N}

, determine the numerical value of

T_B

Intuitive explanation

The fundamental frequency of a vibrating string depends on both its tension and its linear mass density through $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$ . Since both wires have the same frequency despite different lengths and densities, you need to find what tension ratio makes this equation balance—the shorter, denser wire must be under higher tension to compensate. Set the frequency expressions equal and solve for the relationship between $T_A$ and $T_B$ .

Intuitive explanation

Browse the problem library