Question

A transverse wave is described by y(x, t)-(6.50 mm) cos 2π y(r,t) (6.50 mm) cos 2T 28.0cm 0,0360 28.0cm0.0360s Calculate the speed of propagation of this wave in m/s. Sample submission: 1.23

Answer 1

Given that equation of wave,� [y(x, t)] = [(6.5mm) cos 2pi(x/28cm - t/0.036sec)] equation (i)� we know that, general format of wave equation: -� [y(x, t)] = [A cos(kx - wt)] equation (ii)� where (w) = [angular frequency] �= (2pi f)� (k) = [angular wave number]� = (2pi/lambda)� (A) = [amplitude of wave]� Here, on comparing (i), (ii) we have,� (A) = (6.5mm)� (k) = (1/28cm)� (w) = (1/0.036sec) where (w) = [Angular frequency] = (2 pi f) (k) = [angular wave number] = (2 pi/lambda) (A) = [amplitude of wave] \r\n Now,� [speed of propagation(v)] = (w/k)� = [(1/0.036)s/(1/28cm)] = [(28 times 10^-2)m/(0.036sec)] [because (1cm) = (10^-2 m)]� approximatelyequalto (7.8 m/sec)� therefore (v) = (7.8 m/s) is speed of propagation of transverse wave.