A wire of uniform cross-section is stretched between two points 100 cm apart. The wire is fixed at one end…

A wire of uniform cross-section is stretched between two points 100 cm apart. The wire is fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz. (i) What is the velocity of the wave in wire ? (ii) If the weight is reduced to 4 kg, what is the velocity of wave ? What is the wavelength and frequency ?

Solution

Sol. (i) Here, $L=100\ \mathrm{cm}$ and $f_1=750\ \mathrm{Hz}$ \therefore\ $v_1=2Lf_1=2\times100\times750$ $=150000\ \mathrm{cm\,s}^{-1}=1500\ \mathrm{ms}^{-1}$ (ii) : $v_1=\sqrt{T_1/\alpha}$ and $v_2=\sqrt{T_2/\alpha}$ \therefore\ $\dfrac{v_2}{v_1}=\sqrt{\dfrac{T_2}{T_1}}\Rightarrow \dfrac{v_2}{1500}=\sqrt{\dfrac{4}{9}}$ \therefore\ $v_2=1000\ \mathrm{ms}^{-1}$ Wavelength, $\lambda_2=2L=200\ \mathrm{cm}=2\ \mathrm{m}$ Frequency, $f_2=\dfrac{v_2}{\lambda_2}=\dfrac{1000}{2}=500\ \mathrm{Hz}$ Answer: $500\ \mathrm{Hz}$

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