# Cantilever Investigation

1. Dec 14, 2007

### ash1098

[SOLVED] Cantilever Investigation

1.Hi,

I am writing a report on cantilever oscillations, my experiment involves fixing different cantilvers e.g. a ruler to the end of a table then measuring the period and height of oscillations while varing the mass attached to the cantilever, and other varients e.g. lenght of cantilever.

2. I have found these two formulae: (Shown much more clearly in attachments)

T= 2(pi)*[(4ML^3)/(bd^3E)]^1/2

and:

h= 4MgL^3/Ebd^3

where:
b= width of cantilever
d= thickness of cantilever
E= Youngs Modulus
M=Mass
L=Lenght of cantilever
T=period of oscillations
h=height of oscillation

3. I have looked at eqn's involving Hooke's and simple harmonic motion but cannot work out how these formulae have been derived.

Does anyone know how these formulae where derived, or where I can find information on this in general?

Thanks.

ash.

p.s. I have written out the formulae using math open office and attached them in pdf if it helps make them easier to read.

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2. Dec 14, 2007

### rock.freak667

Well I have done a lab to find the Young's Modulus of a loaded cantilever and this is theory which is written down on the paper:

The depression,s,due to a load W(=Mg) at the end of a cantilever of length,l, is

$$s=\frac{Wl^3}{3IE}$$

This strain brings into play internal stresses which produce a restoring force equal to W, i.e. equal to $\frac{3IEs}{l^3}$.

If the acceleration of the load $\frac{d^2s}{dt^2}$ when the cantilever is displaced to produce vertical oscillations,then

$$M\frac{d^2s}{dt^2}=\frac{-3IE}{l^3}s$$

OR
$$\frac{d^2s}{dt^2}+\frac{3IE}{Ml^3}s=0$$

Hence the motion is Simple harmonic and the periodic time,T, is

$$T=2\pi \sqrt{\frac{Ml^3}{3IE}}$$

from which

$$E=\frac{4\pi^2Ml^3}{3IT^2}$$

For a beam of rectangular section:
$$I=\frac{bd^3}{12}$$

I hope that helps in some way

Last edited: Dec 15, 2007
3. Dec 15, 2007

### ash1098

Thanks for your reply rock.freak it's really really useful, just one question:

why does $$T=2\pi \sqrt{\frac{Ml^3}{3IE}}$$

Thanks, ash.

4. Dec 15, 2007

### rock.freak667

Well from
$$M\frac{d^2s}{dt^2}=\frac{-3IE}{l^3}s$$

$$\frac{d^2s}{dt^2}=\frac{-3IE}{Ml^3}s$$

and that is of the form $a=-\omega^2 s$ where $a=\frac{d^2s}{dt^2}$

so from that

$$\omega=\sqrt{\frac{3IE}{Ml^3}}$$

and since it moves with SHM, the period,T, is given by

$$T=\frac{2\pi}{\omega}$$

5. Dec 15, 2007

### ash1098

Oh yeah I see it now, Thanks alot.

Ash.