A string 120 cm in length sustains a standing wave, with the points of string at which the displacement amplitude is equal to `sqrt(2` mm being separated by 5.0 cm, Find the maximum displacement amplitude. Also find the harmonic corresponding to this wave.

Explanation

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