A string `120 cm` in length sustains a standing wave, with the points of the string at which the displacement amplitude is equal to `3.5 mm` being separated by `15.0cm.` Find the maximum displacement amplitude. To which overtone do these oscillations correspond ?

Explanation

We can seefromfigure-6.63, in a stationary wave two successive of equal amplitude, if separated by equal distances then this distance must be `(lambda)/(4)` Thus we have
`(lambda)/(4)` = 15 cm
lamdba = 60 cm

Thus there are four loops in 120 cm length of string. This corresponds of `3^(rd)` overtone oscillations. As shown in figure if we consider origin O is at a node then the amplitude of a general medium particle, at a distance x from O can be given as
`R = A_(0) sin kx`
Where `A_(0) is the maximum displacement amplitude. First point from the origin where amplitude is 3.5 mm is that distance `(lambda)/(8) = 7.5 cm. Thus we have
`3.5 = A_(0) "sin" ((2lambda)/(60) xx 7.5)`
`A_(0) = (3.5)/(sin(pi//4)) = 3.5sqrt(2) mm`