# Derive equation for a cantilever in SHM

I am currently investigating how a hacksaw blade's time period of oscillation changes when I add mass to the end of it or when I change the length it is clamped at.

I found the following equation from an IB worksheet:

$$T^2=\frac{16M\pi^2x^3}{bEd^3}$$

Where:

• $T$ is the time period for one oscillation;
• $M$ is the mass of the cantilever;
• $x$ is the length of the cantilever;
• $b$ is the breadth of the cantilever and $d$ is the thickness of the cantilever;
• and $E$ is the stiffness of the cantilever.

I've looked around the internet and asked my teachers, but I haven't been able to derive this equation from first principles or the equations given in my syllabus.

If anyone knows where this comes from or would like to try and work it out, I would be very greatful.

• Did you look up beam vibrations? – John Alexiou Mar 19 '15 at 18:16
• Can you provide a sketch that indicates the supports and the direction of vibration (side to side, or up and down). – John Alexiou Mar 19 '15 at 18:18
• The cantilever's direction of motion is perpendicular to that of gravity, with the thin, long edge facing towards the ground. – DanielDC88 Mar 22 '15 at 20:27
• Can you please show some kind of sketch. – John Alexiou Mar 23 '15 at 3:13
• – DanielDC88 Mar 24 '15 at 20:17

I was able to derive a similar equation, but not the exact one you mention.

The differential equation of a beam can be stated as

$$E I \frac{\partial^4 u(z,t)}{\partial z^4}+\rho A \frac{\partial^2 u(z,t)}{\partial t^2} =0$$

where: $E$ is the elastic modulus, $I$ is the section area moment, $\rho$ is the mass density, $A$ is the section area and $u(z,t)$ is the deformation as a function of position $z$ and time $t$.

In your case, \begin{aligned} \rho A & = \frac{M}{x} \\ I & = \frac{1}{12} b d^3 \end{aligned}

To fit a static wave of the form $u(z,t)=\sin(\omega t)\sin(n\pi \frac{z}{x})$ you must have $$\left(\frac{\pi^4 E b d^3 n^4}{12 x^4}-\frac{M \omega^2}{x} \right) u(z,t) =0$$ which is solved for $\omega = \frac{2\pi}{T}$ as

$$T^2 = \frac{48 M x^3}{\pi^2 E b d^3 n^4}$$ with the first fundamental frequency when $n=1$. The above is for a simply supported beam, which fits the general shape of the wave $u(z,t)$. For other types of end conditions different wave shapes are needed and different frequencies arise.

• From the sketch I see now this is cantilever beam, and hence the above solution stands for $n=\frac{1}{2}$ or $$T^2 = \frac{768 M x^3}{\pi^2 b E d^3}$$ – John Alexiou Mar 24 '15 at 21:22

The derivation is given in Module 10: Free Vibration of an Undampened 1D Cantilever Beam used by University of Connecticut School of Engineering. The derivation follows the same lines as given in ja74's answer, differing in the boundary conditions (top of p 3).

Note that the formula 10.7 $\gamma_n=\beta_n L=\frac12(2n-1)\pi$ in the UoC document appears to be incorrect, and in equation 10.10 $\beta_n$ should be $\beta_nL=\gamma_n$.

The period of free oscillations of the $n^{th}$ mode can be found from equation 10.10 on p 3 :
$$T_n=\frac{1}{f_n}=\frac{2\pi L^2}{\gamma_n^2}\sqrt{\frac{\rho A}{EI}}$$ in which $\gamma_n$ is a constant depending on mode number $n$. The 1st three mode shapes are illustrated in the following diagram, in which the $Ns$ are nodes :

The lowest mode $(n=1, \gamma_1=1.875)$ applies in your case. From the dimensions of the beam we have $A=bd$ and $I=\frac{1}{12}bd^3$. Making these substitutions we get
$$T_n=\frac{1}{f_n}=6.19\frac{L^2}{d}\sqrt{\frac{\rho }{E}}$$

In your equation we should substitute $M=\rho bdL$ and $x=L$. Then you have
$$T=4\pi \frac{L^2}{d}\sqrt{\frac{\rho }{E}}$$ which has the same form but is a factor of 2.03 larger.

Both equations agree that the relation is $T \propto \frac{L^2}{d}$. The main reason for the discrepancy in your equation is that an extra factor of $x=L$ is hidden within $M=\rho LA$.