# Why does a metal block make a shrill sound but not a wooden block upon hammering?

When hammered, a metal block makes a shrill sound but not a wooden block of identical shape. Is it that the wooden block vibrates with lesser frequency than the metal block? If so, why is that? Also why is the vibration of a metallic block more visible than a wooden block? More amplitude?

The metal block has a relatively low level of internal damping, however the wooden block has a high level of internal damping: Much of the energy imparted to the wooden block is dissipated internally as heat and deformation, also the higher frequencies are damped more than the lower frequencies (it acts like a low pass filter).

So the wooden block will vibrate with fewer harmonics (hence less shrill) and also with lower amplitude (and also with lower duration).

• Hmm. The way humans perceive a note's pitch is by the fundamental frequency of the note. Filtering out more or less higher harmonics doesn't change the pitch, it only changes the timbre. That why a particular note played on a piano and a clarinet sound differently, even if the pitch remains the same. – Gert Mar 1 at 17:51
• Good catch--I edited my answer. – user45664 Mar 1 at 20:22
• Why though: Because of the way the atoms are arranged. And without a crystalline structure (as all metals have), it's not going to sound very good. – Mazura Mar 2 at 1:33
• @Mazura You don't even need a crystalline structure, just any kind of regular structure (at least, I think so?) – somebody Mar 2 at 1:51
• @mithusengupta123 Wood consists of most;y fibers vs metal's molecular structure, and it crushes (permanently deforms, dissipating energy) relatively easily--which presents itself as damping. Whereas metal tends more to compress without dissipating energy--rather converting kinetic energy to potential energy. – user45664 Mar 2 at 16:35

Is it that the wooden block vibrates with lesser frequency than the metal block? If so, why is that?

'Yes', to the first question.

Metal is stiffer than wood and produces higher frequencies (higher pitch).

This follows from the wave equation (here in one dimension):

$$u_{tt}=\frac{E}{\rho}u_{xx}$$

$$E$$ is Young's Modulus and $$\rho$$ the material's density.

When solved, the solution contains a time-dependent factor like this:

$$\cos\Big(\frac{n\pi ct}{L}\Big)$$

where $$n=1,2,3,...$$, and $$c=\sqrt{\frac{E}{\rho}}$$ and $$L$$ a chracteristic length (e.g. the length of a clamped string).

To find the frequencies:

$$\cos \omega t=\cos\Big(\frac{n\pi ct}{L}\Big)$$

$$\omega=2\pi f=\frac{n\pi c}{L}$$

$$f=\frac{n}{2L}\sqrt{\frac{E}{\rho}}$$

The fundamental frequency (for $$n=1$$) is given by:

$$f_1=\frac{1}{2L}\sqrt{\frac{E}{\rho}}$$

So for stiffer materials, i.e. larger $$E$$, the fundamental frequency (as well as the harmonics) is higher.

• Thanks. How will this change if instead of a string I consider a metallic block? Why is the vibration of metal block more visible than the wooden block? Why more amplitude? – mithusengupta123 Mar 1 at 16:05
• The principle that applies to a string also applies to other shapes/objects. The amplitude should mainly depend on how much energy was put in initially, i.e. how hard the block was struck. – Gert Mar 1 at 16:12
• I meant subjected to the same supply of energy i,e., same way of hammering on two blocks of identical shape, one of wood and the other of a metal. – mithusengupta123 Mar 1 at 16:57
• @mithusengupta123 Same reason, frequencies. Higher frequency = shorter wavelength, so for a block of wood, it bends a lot less (also damping as in other answers) – somebody Mar 2 at 1:50