Question

An aluminium wire of length 60 cm is joined to a steel wire of length 80 cm and stretched between two fixed supports. The tension produced is 40 N. Cross-sectional area is 1 mm² (steel) and 3 mm² (aluminium). Minimum frequency of the tuning fork which can produce standing waves with the joint as a node is - (density of Al = 2.6 g cc^–1 and density of steel

= 7.8 g cc^–1) –

• A. 90 Hz
• B. 145 Hz
• C. 180 Hz
• D. 250 Hz

Answer 1

Answer: (C)

180 Hz

Answer 2

f = $\left( \sqrt{\frac{T}{\mu}} \right)$ $\frac{n}{2\mathcal{l}}$ = $\frac{n}{2\mathcal{l}}$ $\sqrt{\frac{T}{(A\rho)}}$

Since frequency remains same

$\frac{n}{2\mathcal{l}_{1}}$ $\sqrt{\frac{T}{A_{1}\rho_{1}}}$= $\frac{p}{2\mathcal{l}_{2}}$ $\sqrt{\frac{T}{A_{2}\rho_{2}}}$

$\frac{n}{p}$ = $\frac{\mathcal{l}_{1}}{\mathcal{l}_{2}}$ Ž $\frac{n}{p}$ = $\frac{4}{3}$

\ f = $\frac { 3 } { 2 \times 0.6 }$ $\sqrt{\frac{40}{10^{–6} \times 2.6 \times 10^{3} \times 3}}$