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 ?

Options

A. `(0.1) sin ( pi//2) x sin (10 pi) t`
B. `(0.1) sin (pi) x sin (10 pi) t`
C. `(0.05) sin (pi//2) x cos (10 pi) t`
D. `(0.04) sin (pi//2) x sin (10 pi) t`

Answer

A

Explanation

`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`