Q. 36

# For the harmonic

Answer :

Given:

The equation of the wave y = 2cos[2π(10t-0.0080x+3.5)]

(a) For the wave separated by a distance of 4 m the equation of vibration will be, or and the phase Comparing this equation to the original equation and the phase the phase difference therefore (b) For a separation of 0.5 m, proceeding as above,  and the phase Comparing this equation to the original equation its phase the phase difference therefore (c) The wavelength of the equation can be found by comparing the equation to the general equation for the wave, we have, k = 0.016π and ω =20π. We know that k is related to the wavelength as where λ is the wavelength. Therefore for k = 0.016π we have For a separation of or 0.625 m,  and the phase Comparing this equation to the original equation its phase the phase difference therefore (d) For a separation of or 0.938 m, proceeding as above  and the phase Comparing this equation to the original equation its phase the phase difference therefore (e) At x = 100 cm the equation of the wave becomes,  We know that ω is and time period T are related as but ω = 20π, At t = T= 0.1 s the equation of the wave is given by, and the phase At t = 5 s the equation of the wave is given by, and the phase the phase difference therefore Rate this question :

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