A simple harmonic wave of amplitude 8 unit travels along positive x-axis. At a given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is + 6 unit and for a particle at a distance of 25 cm from the origin, the displacement is + 4 unit. What is the wavelength (in metre) of the wave?

Step-by-step solution

Y = A sin ⁡ 2 π λ v t - x Y A = sin ⁡ 2 π t T - x λ In first case, Y 1 A = sin ⁡ 2 π t T - x 1 λ Here, Y 1 = + 6 , A = 8 , x 1 = 10 c m 6 8 = sin ⁡ 2 π t T - 10 λ .... (i) Similarly in the second case, 4 8 = sin ⁡ 2 π t T - 25 λ ... (ii) From equation (i), 2π​ t T - 10 λ = sin - 1 ⁡ 6 8 = 0.85 r a d ⇒ t T - 10 λ = 0.14 .... (iii) Similarly from equation (ii), 2 π t T - 25 λ = sin - 1 ⁡ 4 8 = π 6 r a d ⇒ t T = 25 λ +​ 0.08 .... (iv) Subtracting equation (iv) from equation (iii), we get 15 λ = 0.06 ⇒ λ = 0.25 m