JEE Main · 2013 · 07 Apr 2013 · Offline · Waves & Sound

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of

1%

. What is the fundamental frequency of steel if density and elasticity of steel are

7.7 \times 1 0^{3} kg / m^{3}

and

2.2 \times 1 0^{11} N / m^{2}

respectively?

Answer: (A)

Solution

The fundamental of a stretched wire is

f = \frac{1}{2 ℓ} \frac{T}{μ}

, and writing the tension as stress times area and the linear density as

μ = A ρ

lets the area cancel, leaving

f = \frac{1}{2 ℓ} \frac{Y \times strain}{ρ}

. With a

1%

strain,

Y = 2.2 \times 1 0^{11} N/m^{2}

ρ = 7.7 \times 1 0^{3} kg/m^{3}

and

ℓ = 1.5 m

, this gives

f = \frac{1}{3} \frac{2.2 \times 1 0 ^{11} \times 0.01}{7.7 \times 1 0 ^{3}} \approx 178 Hz

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option