# Harmonic wave functions

1. Nov 2, 2004

### triplezero24

Ok heres the deal. The problem gives me this wave function:

y= (15cm) cos[(pi/5cm)x - (pi/12s)t]

I undserstand that 15cm represents the amplitude, 5cm represents the wavelength, and 12s represents the T. What I dont get is that the standard wave function is in this form:

y(x,t)= A cos[(2pi/lambda)x - (2pi/T)t]

I thought I was good at algebra, but I can't seem to figure out how to the the pi's in the given equation to be at a factor of 2 with the rest of it being the same. The problem asks me to find: amplitude (which I got 15cm), the wavelength (I got 2.25cm), the T (I got 6s), and speed and direction. Please help if you can!

Thanks,
Eric

2. Nov 2, 2004

### spacetime

y= (15cm) cos[(pi/5cm)x - (pi/12s)t]

can be written as

y= (15cm) cos[(2*pi/10cm)x - (2*pi/24s)t]

So, the wavelength is 10 cm and T is 24s.

Speed is 10/24 cm/s and direction is positive x.

3. Nov 2, 2004

### Pyrrhus

Or use Partial derivatives for speed.

$$\frac{\partial y}{\partial t} = v$$

Or

$$v = f \lambda = \frac{\omega}{k}$$

Last edited: Nov 2, 2004