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A string 120 cmin length sustains a standing wave, with the points ofstring at which the displacement amplitude is equal to sqrt2mm being separated by 15.0 cm, Find the maximum displacement amplitude. Also find the harmonic corresponding to this wave. |
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Answer» Solution :From figure. Point A, B, C, D and E are having equal displacement amplitude. Further, `x_E - x_A = lambda = 4 xx15 = 60 cm`  As `lambda =(21)/n = (2 xx 120)/n =60` `:.n = (2 xx 120)/60 =4` So, it corresponds to `4^(th)` harmonic. Also,distance of node from A is 7.5 cm and no node is between the.Taking node at ORIGIN, the amplitude of STATIONARY wave can be written as, a = A sin kx Here `a = sqrt2 mm, k = (2pi)/lambda =(2pi)/60 and x =7.5 cm` `:. sqrt2 = A sin ((2pi)/(60) xx 7.5)= A sin ""(pi)/4` Hence, A = 2 mm |
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