A wire of density `9 xx 10^(3) kg//m^(3)` is stretched between two clamps 1 m apart and is subjected to an extension of `4.9 xx 10^(-4) m`. The lowest frequency of transverse vibration in the wire is `(Y = 9 xx 10^(10) N//m^(2))`

Options

A. 38 Hz
B. 36 Hz
C. 35 Hz
D. 32 Hz

Answer

B

Explanation

The lowest frequency (fundamental frequency) of the transverse vibrations in the wire is `upsilon=(1)/(2L)sqrt((T)/(mu))" "…(i)`
where T is the tension, L is the length and `mu` is the mass per unit of the wire. As `mu=rhoA" "…(ii)`
where `rho` is the density and A is the area of cross-section of the wire.
Young.s modulus, `Y=(TL)/(ADeltaL)orT=(YADeltaL)/(L)" "...(iii)`
Divide (iii) by (ii), we get `(T)/(mu)=(YDeltaL)/(Lrho)`
Substituting this value in equation (i), we get `upsilon=(1)/(2L)sqrt((YDeltaL)/(rhoL))`
Substituting the given values, we get
`upsilon=(1)/(2xx1)sqrt((9xx10^(10)xx4.9xx10^(-4))/(9xx10^(3)xx1))=(70)/(2)=35Hz`