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A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus. |
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Answer» Correct Answer - 1.985xx10^11N/m^2 `L=40cm=.4m` mass=3.2g=3.2x10^-3kg` `:. Mass per unit length `m=(3.2xx10^-3)/0.4=8xx10^-3(kg)/m` Change in length `/_\L=40.05=80xx10^-3(kg)/m` Strain` =(/_\L)/L=((0.05xx10^-2))/0.4` `=0.125xx10^-2` `f=220Hz` `f=1/(2L)sqrt(T/m)` `=1/(2xx(0.4005))` `sqrt((T/(8xx10^-3)))` `rarr 220xx220=[1/((0.801)^2)]xxTxx{10^3/8)` `rarr Txx1000=220xx220xx0.641xx0.8` `rarr T=248.19N` Strain =`248.19/1(1mm^2)` Strain =`248.19/(1mm^2)` `=248.19/10^-6=248.19xx10^6` `Y`=stress/strain `=((248.19xx10^6))/((0.125xx10^-2)) =1985.52xx10^8 =1.985xx10^11N/m^2` |
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