Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the +x-axis and is fixed at x=0.
(a) find the displacement of a point on the string as a function of position and time.
(b) Find the speed of propagation of a transverse wave in the string.
(c) Find the amplitude at a point 3.0 cm to the right of an antinode.

Answer

A, B, C, D

Explanation

(a) `lambda/2 = 15 cm`
`:. lambda= 30 cm`
`k = (2pi)/lambda = pi/15 cm^-1`
`omega = (2pi)/T = (2pi)/0.075 s^-1`
Since, x=0 is a node we will write sin
equation.
`:. A_x = A_(max) sin kx` …….(i)
and `y= A_x sin omegat`
(b) ` v = omega/k = ((2pi//0.075))/((pi//15))`
`= 400 cm//s = 4m//s.`
(c) `x=0`

`x = 7.5 - 3 = 4.5 cm `
Form Eq. (i),
`A_x = (0.85 cm ) sin (pi/15xx 4.5)`
=0.688 cm .