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Asked by sarveshvibrantacademy | 04 Apr, 2019, 12:04: PM
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For the symmetrical triangular function of maximum height 0.4m  at a distance x=0.5 m, functional form of displacement y is given by
y = 0.8x,  when 0 ≤ x ≤ 0.5
y = 0.8 - 0.8x  when 0.5≤ x ≤ 1.0
y = 0 when x ≥ 1.0

Wavefunction y(x,t) of the travelling pulse is obtained by substituting x = (x - 24t)  because pulse travelling velocity is 24 m/s.
Wave function of the travelling pulse is given by
 
y(x,t)= 0.8(x-24t)               0 ≤ (x-24t) ≤ 0.5
        = 0.8 - 0.8(x-24t)      0.5≤ (x-24t) ≤ 1.0
        = 0                             (x-24t) ≥ 1.0
 
In the above wave function,  t is in second, x and y are in metre.
 
velocity of particle dy/dt is obtained by differentiating above wave function y(x,t) with respect to time.
 
v(x,t) = dy(x,t)/dt  = - 19.2   0 ≤ (x-24t) ≤ 0.5
                             =  +19.2  0.5≤ (x-24t) ≤ 1.0
 
In the velocity function given above,  v is in m/s

To get velocity as a  function of time at x=1.0 m, we change the space coordinate of above eqn. into time coordinate by substituting x = 1
 
v(1,t) = 19.2 ,      0≤ t ≤ (1/48)
          = -19.2 ,     (1/48) ≤ t ≤ (1/24)
          = 0                t ≥ (1/24)
 
Above velocity function is plotted in figure
Answered by Thiyagarajan K | 05 Apr, 2019, 10:34: AM

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