# Bulk Modulus formula help

#### AakashPandita

bulk modulus = Δp/ΔV/V

Why do we need to divide change in volume by the original volume?

why is it important?

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#### Spinnor

Gold Member
Say you have a spring with spring constant k that is one inch tall with no weight on it. Now say you put on top the spring a mass which compresses the spring 1/10 an inch. Now stack two such springs on top of each other and place the same mass on top of the stacked springs. It should not be too hard to convince yourself that the two springs together will now compress by 2/10 an inch. The compressive properties of the springs, k, has not changed but with more springs you get more movement. Compressive material acts like the springs, more springs stacked on top of one another you get more movement for a given force.

If we apply compressive forces to some object we want to know how much it will change shape but that depends on the size and its properties.

Does this make sense or help?

#### Bill_K

Because if we used just Δp/ΔV, the bulk modulus would depend on the size of the sample. We want a quantity that is independent of the sample size.

#### AakashPandita

say i have 2 iron blocks. 1 is a small block and other is huge.
I need to decrease each of their volumes by 10 m^3.
would the force required in each case be the same or not?

#### AakashPandita

say i have 2 iron blocks. 1 is a small block and other is huge.
I need to decrease each of their volumes by 10 m^3.
would the force required in each case be the same or not?

#### nasu

No, it would not.
What if the small block is 1 m^3 and the large one is 100 m^3?
Assume you find the force you need to decrease the volume of the large one by 1 m^3.
Will the same force decrease the volume of the first one by 10m^3? Will any force be able to do this?

Can you see what kind of problems will arise trying to define a property in terms of absolute values? Not that is not possible.
Absolute value of volume change is not a property of the material but of the specific sample.

#### AakashPandita

this is interesting! thanks.

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