# Homework Help: Amplitude of particle in SHM

1. Nov 11, 2007

### Ginerva123

1. The problem statement, all variables and given/known data
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.

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

3. 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.

2. Nov 11, 2007

### learningphysics

What is the equation of a standing wave?

why are you trying to get the second derivative?

3. Nov 12, 2007

### 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.

4. Nov 12, 2007

### learningphysics

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 ?

Last edited: Nov 12, 2007
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