The wave function for the wave pulse is `y(x,t)=(0.1C^(3))/(C^(2)+(x-vt)^(2))` with C=4cm at x=0 the displacement y(x,t) is observed to decrease from its maximum value to half of its maximum value in time `t=2times10^(-3)s`.

The wave function of a pulse is given by y=(5)/((4x+6t)^(2)) where x and y are in metre and t is in second :- (i) Identify the direction of propagation (ii) Determine the wave velocity of the pulse

A simple harmonic oscillator at the point x=0 genrates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. the rope has a linear mass density of 50.0 g//m and is strectches with a tension of 5.00 N. (a) determine the speed of the wave. (b) find the wavelength. (c ) write the wave function y(x,t) for the wave, Assume that the oscillator has its maximum upward displacement at time t=0. (d) find the maximum transverse acceleration at time t=0. (d) find the maximum transverse of points on the rope. (e) in the discussion of transverse waves in this chapter, the force of gravity was ignored. is that a reasonable assumption for this wave? explain.

A travelling wave pulse is given by y=(4)/(3x^(2)+48t^(2)+24xt+2) where x and y are in metre and t is in second. The velocity of wave is :-

At time t=0 , y(x) equation of a wave pulse is y=(10)/(2+(x-4)^(2)) and at t=2s , y(x) equation of the same wave pulse is y=(10)/(2+(x+4)^(2)) Here, y is in mm and x in metres. Find the wave velocity.

The particles position as a function of time is given as x=t^(3)-3t^(2)+6, then maximum value of x occurs when "t=