In a standing wave experiment , a 1.2 - kg horizontal rope is fixed in place at its two ends ( x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode , at frequency of 5.0 Hz. At t = 0 , the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y - axis with a transverse velocity 3.14 m//s. What is the correct expression of the standing wave equation ?

In a standing wave experiment , a 1.2 - kg horizontal rope is fixed in

1.	In a standing wave experiment , a 1.2 - kg horizontal rope is fixed in place at its two ends ( x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode , at frequency of 5.0 Hz. At t = 0 , the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y - axis with a transverse velocity 3.14 m//s. What is the correct expression of the standing wave equation ?
Answer» `(0.1) sin ( pi//2) x sin (10 pi) t` `(0.1) sin (pi) x sin (10 pi) t` `(0.05) sin (pi//2) x cos (10 pi) t` `(0.04) sin (pi//2) x sin (10 pi) t` Solution :`mu = (1.2)/(2) = 0.6 kg//m` `N = 5 Hz` `lambda = 2 l = 4 m` `V = n lambda = 5 XX 4 = 20 m//s` Using `v = sqrt((T)/(mu))` `T = 20^(2) xx 0.6 = 240 N` `((delta y)/( delta t))_(max) = 3.14 m//s` `( 2 A)omega = 3.14` Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` EQUATION of standing wave is `y = (0.1) sin (pi)/(2) s sin (10 pi) t`