# Amplitude of particle in SHM

Ginerva123

## 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.

Homework Helper
What is the equation of a standing wave?

why are you trying to get the second derivative?

Ginerva123
y(x,t)=(Asinkx)sinwt
I thought the second derivative wrt x was equal to the second derivative wrt t by the inverse of v squared. Or maybe not.

Homework Helper
y(x,t)=(Asinkx)sinwt

Yes, that's the right equation.

I still don't understand... why do you need the second derivative with respect to x, or with respect to time? you need the amplitude at a certain x value right?

you've got the equation of the standing wave.

y(x,t)=(Asinkx)sinwt

The amplitude at any x value is simply Asinkx... because the maximum value that sinwt takes over time is 1... ie amplitude at x is Asinkx(1) = Asinkx.

you have A. you can get k ... You need the amplitude at a point 0.03m to the right of an anti-node... so just choose any anti-node... you've got the wavelength = 0.30m. where is the first anti-node?

what do you get for Asinkx, where x = 0.03 + position of anti-node ?

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