# Resonance prac question (1 Viewer)

#### erucibon

##### New Member
In the prac, different tuning forks were placed over a pipe in a measuring cylinder of water. When the 128Hz tuning fork was used, the resonance point couldn't be heard. Not quite sure why this is the case, any suggestions are appreciated.
thanks

#### Drdusk

##### sin(x) = tan(x) = x
In the prac, different tuning forks were placed over a pipe in a measuring cylinder of water. When the 128Hz tuning fork was used, the resonance point couldn't be heard. Not quite sure why this is the case, any suggestions are appreciated.
thanks
Resonance is essentially a standing wave. In order for resonance to occur the Frequency f must follow the rule below:

$f = \frac{nv}{2L} \text{where}\hspace{2mm} n \in \mathbb{Z}$

where L is $L_{P} - L_{S}$ where $L_S$ is the submerged length, i.e. just the amount of pipe that is submerged under water and $L_P$ is the total length of the pipe. This gives us:

$f = \frac{nv}{2(L_{P} - L_{S})}, n \in \mathbb{Z}$

$n = \frac{2f(L_P - L_S)}{v}, n \in \mathbb{Z}$

Now if you measure the length of the pipe, and the amount submerged etc. You can manipulate that expression to calculate n. If n IS NOT an INTEGER, then obviously resonance will not be observed, because the pipe simply does not have the correct length for a proper standing wave to form.

NOTE: This is done assuming end effects are negligible.