- #1

Ginerva123

- 14

- 0

## Homework Statement

Adjacent antinodes of a standing wave on a string are a distance 15.0cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850cm and period 0.0750s. The string lies along the +x - axis and is fixed at x=0.

At time t=0, all points on the string are at their minimum displacement.

Find the amplitude at a point a distance 3.0cm to the right of an antinode.

## Homework Equations

d^y(x,t) /dx^2 = 1/v^2 . d^2y(x,t)/(dt^2)

## The Attempt at a Solution

wavlength is 2 x 15 cm = .30m

f is 1/T = 13.333 Hz

so v = 4 m/s, v^2 = 16 and 1/v^2 is 0.0625

However, I' having trouble getting the second derivative of the wave equation using the point 3.0 cm to the right of the antinode.