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A simple harmonic wave of amplitude `8` units travels along positive x-axis. At any given instant of time, for a particle at a distance of `10 cm` from the origin, the displacement is `+ 6 units`, and for a particle at a distance of `25 cm` from the origin, the displacement is `+ 4` units. Calculate the wavelength. |
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Answer» Correct Answer - A::B::C::D `y =A sin (2pi)/(lambda)(vt - x)` or `(y)/(A) = sin 2pi ((t)/(T) -(x)/(lambda))` In the first case, `y_(1) = + 6 , A = 8 , x_(1) = 10cm` `(6)/(8) = sin 2pi((t)/(T) - (x_(1))/(lambda))` Here, `y_(1) = + 6 , A = 8 , x_(1) =10 cm` `:. (6)/(8) = sin 2pi ((t)/(T) - (10)/(lambda))` ...(i) Similarly in the second case, `(4)/(8) = sin 2pi((t)/(T) - (25)/(lambda))` ...(ii) From Eq. (i), `2pi((t)/(T) - (10)/(lambda)) = sin^(-1) ((6)/(8)) = 0.85 rad` or `(t)/(T)-(10)/(lambda) = 0.14` ...(iii) Similarly from Eq. `(ii)`, `2pi((t)/(T) - (25)/(lambda)) = sin^(-1) ((4)/(8))= (pi)/(6) rad` or `(t)/(T) - (25)/(lambda) = 0.08` ...(iv) Substracting Eq. (iv) from Eq. (iii), we get `(15)/(lambda) = 0.06` `:. lambda =250 cm` |
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