(i) The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. Find the…
(i) The length of the wire shown in figure between the pulleys is 1.5 m and its mass is 12.0 g. Find the frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulleys at rest.
(ii) A string fixed at both ends, vibrates in a resonant mode with a separation of 2.0 cm between the consecutive nodes. For the next higher resonant frequency, this separation is reduced to 1.6 cm. Find the length of the string.
Solution
Sol. (i) Given, L = 1.5 m, M = 12 g
(Tension shown by two 9 kg hanging masses)
$T = 90\ \mathrm{N}$ (Acting downward)
$\therefore\ \alpha = \dfrac{M}{L} = \dfrac{12 \times 10^{-3}}{1.5} = 8 \times 10^{-3}\ \mathrm{kgm^{-1}}$
As, $f = \dfrac{2v}{2L}\;(\therefore\ n = 2)$
$f = \dfrac{2}{2L} \sqrt{\dfrac{T}{\alpha}} = \dfrac{1}{1.5} \sqrt{\dfrac{90}{8 \times 10^{-3}}} = 70.71\ \mathrm{Hz}$
(ii) For $k$th harmonic, $k\cdot \dfrac{\lambda}{2} = L \Rightarrow k(2) = L$ \dots (i)
For $(k+1)$th harmonic,
$(k+1)\dfrac{\lambda'}{2} = L \Rightarrow (k+1)(1.6) = L$ \dots (ii)
From Eqs. (i) and (ii), $2k = 1.6(k+1)$
$0.4\,k = 1.6 \Rightarrow k = 4$
Hence, $L = 2k = 2 \times 4 = 8\ \mathrm{cm}$
Answer: 8 cm