First slide

Stationary waves

Question

A wire of density 9 × $10^{- 3}$ kg $c m^{- 3}$ is stretched between two clamps 1m apart. The resulting strain in the wire is 4.9 × $10^{- 4} .$ The lowest frequency of the transverse vibrations in the wire is (Young’s modulus of wire Y = 9 × $10^{10} N m^{- 2}$ ), (to the nearest integer), ______

Moderate

Solution

Lowest frequency is fundamental frequency
$\begin{array}{l} f_{0} = \frac{1}{2 l} \sqrt{\frac{T}{μ}} [Y = \frac{T l}{A Δ l}] \\ \Rightarrow f_{0} = \frac{1}{2 l} \sqrt{\frac{T}{ρ A}} [\frac{T}{A} = \frac{Y Δ l}{l}] \\ \Rightarrow f_{0} = \frac{1}{2 l} \sqrt{\frac{Y Δ l}{ρ l}} \\ G i v e n : l = 1; \frac{Δ l}{l} = 4.9 \times 10^{- 4}; Y = 9 \times 10^{10}; ρ = 9 \times 10^{+ 3} \frac{k g}{m^{3}} \\ S o \Rightarrow f_{0} = \frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{- 4}}{9 \times 10^{3}}} = \frac{1}{2} \sqrt{4.9 \times 10^{3}} = \frac{1}{2} \sqrt{49 \times 10^{2}} = \frac{70}{2} = 35 H z \end{array}$

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