In the figure given below is shown displacement-distance (y – x) graph of a transverse wave travelling along a string at the instant t = 0.25 s. At the instant t = 0, the end A of the string was in the mean position. Find out: (i) Frequency, wave-length and velocity of the wave, (ii) Equation of the wave, (iii) Displacement of B at the instant t = 1 second.
100 cm/s, , –10cm.
(i) According to the graph, in the time-interval from t = 0 to t = 0.25 second, the end A of the string reaches from the mean position to the maximum-displacement position. This means that one-fourth of an oscillation is completed in 0.25 second.
From the graph: wavelength, λ = 125 cm – 25 cm = 100 cm.
(ii) = 100 cm in the equation
Now, from graph, at t = 0.25 s, we have y = –10 cm at x = 0, that is, at point A. Thus
Substituting this value of a in equation (i), we get
This is the wave equation.
(iii) To find the displacement of point B at the instant t = 1s, we substitutet = 1s and x = 75 cm in the above equation:
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