InterviewSolution
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InterviewSolution
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A travelling harmonic wave on a string is described by y(x, t) = 7.5 sin (0.005 x +12t + π /4) 1. what are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation? 2. Locate the points of the string which have the same transverse displacements and velocity as the x a 1 cm point at t = 2 s, 5 s, and 11s. |
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Answer» Given: y (x, t) = 7.5 sin (0.005 x + 12t + π /4) 1. At x=1 cm and t = 1s y (1, 1)= 7.5 Sin(0.005 +12 + π /4) = 7.5 sin (12.005 + π /4) = 1.67 cm Velocity of oscillation : v = \(\frac{d(Y (x,t))}{dt}\) = d/dt (7.5 sin (0.005 x + 12t + π/4) dt = 7.5 × 12 cos (0.005 x+ 12t + π/4) At x = 1 cm and t = 1 s v = 7.5 × 12 cos(0.005 + 12 + π/4) = 87.75 cm s-1 We know that velocity of wave propagation = w/k Here w = 12 s-1 and k = 0.005 cm-1 ∴ Velocity of wave propagation = \(\frac{12s^{-1}}{0.005cm^{-1}}\) = 24 ms-1 ∴ At x = 1 cm and t = 1 s velocity of oscillation is not equal to velocity of wave propagation. 2. In a wave, all the points which are separated by a distance ± λ, ±2λ ……..from x = 1 cm will have same transverse displacements and velocity. For the given wave , λ= \(\frac{2π}{0.005}\) = ±12.56 cm, +25.12 m….From x = 1 cm will have the same displacements and velocity as at x = 1 cm, t = 2s, 5s and 11 s. |
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