Question

a steel wire hangs vertically from a fixed point supporting a weight of 80n at its lower end, the diameter of wire is 0.50mm and it's length form the fixed point to the weight is 1.5m. calculate fundamental frequency emitted by wire when it's pluncked​

Answer 1

Given info : a steel wire hangs vertically from a fixed point supporting a weight of 80N at its lower end, the diameter of wire is 0.50mm and it's length from the fixed point to the weight is 1.5m.

To find : the fundamental frequency emitted by the wire when it is plunked

solution : tension acting on the wire is T = 80 N

length of wire from the fixed point to the weight is L = 1.5 m

we know, density of steel is ρ = 7800 kg/m³

so the linear mass density of steel wire, μ = ρ A

= 7800 kg/m³ × π(0.50 × 10⁻³ m)²/4 m² = 1.53 × 10⁻³ kg/m

now the fundamental frequency emitted by the wire is f = $\frac{1}{2L}\sqrt{\frac{T}{\mu} }$

       =  $\frac{1}{2\times1.5} \sqrt{\frac{80}{1.53\times10^{-3}} }$

       = 76 Hz

therefore the fundamental frequency emitted by the wire when it is plunked is 76 Hz.

Answer 2

Answer:

76Hz

Explanation:

Solved by the formula f = 1/2L × √T/μ

μ = PA

P = Density

A = Area = 2πr

r = Radius = (D/2)²

D = Diameter