putting \(n = 340Hz, v = 340 m/s,\) the length of air column in the pipe can be
\(l = \frac{{(2N - 1)340}}{{4 \times 340}} = \frac{{(2N - 1)}}{4}m = \frac{{(2N - 1) \times 100}}{4}cm\)
For \(N = 1, 2, 3, ... \) we get \(l = 25 cm, 75 cm, 125 cm ...\)
As the tube is only \(120 cm\) long, length of air column after water is poured in it may be \(25 cm\) or \(75 cm\) only, \(125 cm\) is not possible, the corresponding length of water column in the tube will be \((120 -25) cm = 95 cm\) or \((120 -75) cm = 45 cm. \)
Thus minimum length of water column is \(45 cm.\)
${y_1} = 0.05\,\cos \,\left( {0.50\,\pi x - 100\,\pi t} \right)$
${y_2} = 0.05\,\cos \,\left( {0.46\,\pi x - 92\,\pi t} \right)$
તો તેનો વેગ $m/s$માં કેટલો મળે?