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?

Explanation

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`