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 5) A sample of Granite is observed to have a P- wave velocity of 5 km/sec and density of 2600 kg/m3. The Bulk modules of the granite, assuming it to be a poissions solid is ____________ kilo-Pascal(Kpa).

(Thanks to Chandrasekhar, ANU)


Solution:
Given that V_{p}= 5 km/sec =5000m/sec
Density(ρ) = 2600 kg/m3
Bulk Modules(K)= ?
Poisson solid (\nu) =0.25 

\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(1-\upsilon)}{(1-2\upsilon)}}

\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(1-0.25)}{(1-2*0.25)}}

\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(0.75)}{(1-0.5)}}

\frac{V_{p}}{V_{s}} =\sqrt{\frac{1.5}{0.5}}

\frac{V_{p}}{V_{s}} =\sqrt{\frac{1.5}{0.5}}

\frac{V_{p}}{V_{s}} =\sqrt{3}
  
V_{s}= \frac{V_{p}}{\sqrt{3}}
V_{s}= \frac{5000}{1.732}
V_{s}= 2886.83 m/sec^2
We know that velocity of S-wave can be defined as follws


V_{s}^{2}=\frac{\mu}{\rho}
\mu=\rho\times V_{s}^{2}
\mu=(2600)\times (2886.83)^{2}
\mu = 2.16\times 10^{10}


Velocity of the P-wave can be defined as follows
V_{p}^{2}= \frac{K + \frac{4}{3}\mu}{\rho}
(5000)^{2}= \frac{K + \frac{4}{3}( 2.16\times 10^{10})}{2600}
6.5\times 10^{10} = K + 2.88 \times 10^{10}
K = 3.62\times 10^{10} pascal
K= 3.62 \times 10^{7} kilo - pascal
K= 36200000 Kpa


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