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Frequency of SONAR =40 kHz Speed of enemy submarine vo=360 km h-1 = 100 m s-1 This is the frequency which would be observed and reflected by the enemy submarine but won't appear the same to the SONAR(source) as again there is relative motion between the source(enemy submarine) and the observer(SONAR) The frequency which would be received by the SONAR is
A=0.05 m Tension in the string is T=mg The speed of the wave in the string is v Angular frequency of the wave is Since at t=0, the left end (fork end) of the string x=0 has zero transverse displacement (y=0) and is moving along the positive y-direction, the initial phase is zero.  Taking the left to the right direction as positive we have Here t is in seconds and x and y are in metres.
(a) The pulse does not have a definite frequency or wavelength however the wave has definite speed given the medium is non-dispersive. (b) The frequency of the note produced by the whistle is not 0.05 Hz. It only implies the frequency of repetition of the pip of the whistle is 0.05 Hz,
The wavelength of the given wave is The points with the same displacements and velocity at the same instant of time are separated by distances . The points of the string which have the same transverse displacements and velocity as the  point at 5 s and 11 s would be at a distance of  from x = 1cm. Therefore all points at distances  from the point x=1cm would have the same transverse...
The displacement of oscillation of a point at x = 1 cm and t = 1 s is The general expression for the velocity of oscillation is k=0.005 cm-1 The velocity of propagation of the wave is The velocity of oscillation of point at x = 1 cm and t = 1 cm is not equal to the propagation of the wave.