# Problem Involving Exponentials (Sound)

1. Mar 27, 2006

### richnfg

I'm doing a piece of A-level physics coursework at school at the moment and seens I do Maths as well this shouldn't be hard for me, but I just can't get my head around it. So, any help would be really apprieciated!

For my coursework, I'm testing different materials for sound-proofing and then looking at the structure of the material - seeing which works best and why.

One side I have a signal generator with a speaker set a set distance away from a microphone and an oscilloscope (that measures the potential difference, or the voltage). In between the gap I insert materials and measure the change in the voltage (for example, with just air in the gap I recieved a reading of 20mv, when cardboard was inserted it dropped to 10mv).

After my first practical I realised a problem...I didn't even think to take into account the thickness of the material, and they were all of different thicknesses! I spoke to my Teacher about this and he came up with the idea of using an equation using logs to find a constant that would over come this problem.

He said:

He said that plotting the Voltage (from the material) against thickness would result in an exponential curve and hence you could state:

Vm = Vo x e^(-ct)

Where Vm is the voltage from a material (the reading), Vo is in air, c is some unknown constant and t is the thickness.

He then took logs (simple enough):

lnVm = lnVo - ct
lnVo = lnVm + ct

Then apparently you can plot lnV against t to get a gradient, which is C.

Could someone please explain how you get from that final equation to plotting that graph? And also how I would reapply the answer with the constant to my old data to form the new more exact data.

Thanks, Rich.

2. Mar 27, 2006

### Kurdt

Staff Emeritus
The reason logs are used is to produce a straight line graph in which the radient is constant. This is done to make it easier to find c. So you'd plot the natural log of frequency against t and work out the gradient to find c and you can stick it back in the original equation.

3. Mar 27, 2006

### HallsofIvy

Staff Emeritus
First, you don't go from "that final equation to plotting that graph"! You go the other way. Do exactly what your teacher suggested: plot log of voltage against time. Assuming your basic idea is correct, that the function really is exponential, you should get approximately a straight line. You can calculate the slope of the line and get its equation, then work back to the exponential.

4. Mar 27, 2006

### richnfg

Yeah, I understand and get all that.

Just need to know how I would then apply this back to get my new set of results were every material has the same thickness.

5. Mar 28, 2006