Question

A 40 cm wire having a mass of 3⋅2 g is stretched between two fixed supports 40⋅05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is 1⋅0 mm2, find its Young modulus.

Answer 1

Given:

Length of the wire (L) = 40 cm = 0.40 m

Mass of the wire = 3.2 g = 0.003 kg

Distance between the two fixed supports of the wire = 40.05 cm

Fundamental mode frequency = 220 Hz

Therefore, linear mass density of the wire (m) is given by:

m=0.00320.4=8×10−3 kg/mChange in length, ∆L=40.05−40=0.05 cm       =0.05×10−2 mStrain=∆LL=0.05×10−20.4           =0.125×10−2f0=12LTm‾‾√    =12×(0.4005) T8×10−3‾‾‾‾‾‾√

⇒220×220=[1(0.801)2]×T×(1038)⇒T×1000=220×220×0.641×0.8⇒T=248.19 NStress=TensionArea=248.191 mm2=248.1910−6

Answer 2

Answer:

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