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The length of a sonometer wire is `0.75m`, and density `9xx10^(3)m`. It can bear a stress of `8.1xx10^(8)N//m^(2)` without exceeding the elastice limit. What is the fundamental frequency that can be produced in the wire? |
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Answer» Here, `l=0.75m, rho=9xx10^(3)kg//m^(3)`. stress`=8.1xx10^(8)N//m^(2),v=?` Let a be the area of cross section of the wire. If m is mass of unit length of wire, then `m=axx1xxrho` `v=(1)/(2l)sqrt((T)/(m))=(1)/(2l)sqrt((stressxxa)/(axx1xxrho))` `=(1)/(2xx0.75)sqrt((8.1xx10^(8))/(9xx10^(3)))` `v=(1)/(1.5)xx3xx10^(2)=200Hz` |
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