Question

A wire 1m long and having a density of 9x10^3 kg/m^3 is stretched by 4.9x10^-4 m and tied between two rigid supports .if Y=9x10^10 N/m^2 ,the minimum frequency of transverse vibrations in the string will be ....plz explain i will definitely mark it as a brainliest ans if it ans will be correct ​

Answer 1

length of wire, l = 1m

density, d = 9 × 10³ kg/m³

extension of wire, ∆l = 4 × 10^-4 m

Let T is tension in wire.

we know, Young's modulus = stress/strain

Y = Tl/(A × ∆l)

9 × 10^10 N/m² = T × 1m/(A × 4 × 10^-4 m)

or, 9 × 10^10 = (T/A) × (1/4) × 10⁴

or, 36 × 10^6 = (T/A)

or, T = 36 × 10^6 A ......(1)

here , A is cross sectional area of wire

density = mass / volume

mass = density × volume

m, = d × (Al) ......(2) [ volume of wire is cross sectional area × length of wire ]

minimum frequency, f = 1/2l√{T/m}

= 1/2l √{T/(dAl)} [ from equation (2), ]

= 1/2l √{36 × 10^6 A /(dAl)}

= 1/2l × 6 × 10³ √{1/dl}

now putting, l = 1m, d = 9 × 10³

= 3 × 10³ × √{1/9 × 10³}

= 10³ × 1/10√(10)

= 10²/√10

= 10√10

≈ 316.2 Hz