Question

A wire pf density 9*10^3 kg/m^3 is stretched between 2 clamps 1m apart & is subjected to an extension of 4.9 * 10^-4 m . What will b the lowest frequency of vibration in wire?
(1) 25 Hz (2) 35Hz (3) 45 Hz (4) 55 Hz

Answer 1

Standing waves, with tension in the wire stretched.. there are nodes at the ends.

I think the information on the Young's modulus or some other property is missing...

Density = 9000  kg/m³         μ = mass per unit length
L = 1 meter                    ΔL = 4.9 * 10⁻⁴  meter = 0.00049 m

Y = Stress / strain  =  Tension * L / (A * ΔL)
Y = T / A  * 1/0.00049 =   2040.816 T/A

d = mass / volume = mass / (L * A)  =  μ / A  = 9000 kg/m³
       =>  A = μ/9000 m²

T =  Y A / 2040.816 =  Y μ / 1836734
   T / μ  =  5.44 * 10⁻⁷  Y

lowest Frequency of vibration = fundamental frequency = f₀
wavelength = λ;  There is only one loop in the vibrating string.. tied (fixed) at both ends.
             L = λ / 2      =>    λ = 2 meters

  velocity of the transverse wave on a stretched string tied at both ends :
                v = √ (T / μ)  =   λ * f₀

    f₀ = v / λ = 3.689 * 10⁻⁴ √Y
           where Y = Young's modulus of the material of the wire.

If  Y = 9 * 10¹⁰ Newtons/meter²  then we get :
         f₀ = 110.67 Hz