JEE Challenger
More from Mechanical Properties of Solids
PhyMedAdv 2010 • Paper 1

Oscillation of Suspended Mass on Elastic Wire to Find Young Modulus

A 0.1 kg0.1\text{ kg} mass is suspended from a wire of negligible mass. The length of the wire is 1 m1\text{ m} and its cross-sectional area is 4.9×10−7 m24.9\times 10^{-7}\text{ m}^2. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad s−1140\text{ rad s}^{-1}. If the Young's modulus of the material of the wire is n×109 Nm−2n\times 10^9\text{ Nm}^{-2}, the value of nn is

Official Numerical Answer4

Topics & Concepts

Mechanical Properties of SolidsElasticitySHM

Step-by-Step Solution

To find the Young's modulus of the material of the wire, we can model the wire as an ideal spring.

The equivalent spring constant kk of a wire of length LL, cross-sectional area AA, and Young's modulus YY is given by:
k=YALk = \frac{YA}{L}

When a mass mm suspended from the wire is displaced vertically, it executes Simple Harmonic Motion (SHM). The angular frequency ω\omega of these vertical oscillations is:
ω=km=YAmL\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{YA}{mL}}

Squaring both sides and solving for YY:
Y=mLω2AY = \frac{m L \omega^2}{A}

Given:

  • Mass, m=0.1 kgm = 0.1\text{ kg}
  • Length, L=1 mL = 1\text{ m}
  • Area of cross-section, A=4.9×10−7 m2A = 4.9 \times 10^{-7}\text{ m}^2
  • Angular frequency, ω=140 rad s−1\omega = 140\text{ rad s}^{-1}

Substitute these values into the formula:
Y=0.1×1×(140)24.9×10−7Y = \frac{0.1 \times 1 \times (140)^2}{4.9 \times 10^{-7}}
Y=0.1×196004.9×10−7=19604.9×10−7=400×107 N m−2Y = \frac{0.1 \times 19600}{4.9 \times 10^{-7}} = \frac{1960}{4.9 \times 10^{-7}} = 400 \times 10^7\text{ N m}^{-2}
Y=4×109 N m−2Y = 4 \times 10^9\text{ N m}^{-2}

Comparing this with Y=n×109 N m−2Y = n \times 10^9\text{ N m}^{-2}, we get:
n=4n = 4

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