Question

# A 40 cm long wire having a mass 3.2 gm and area of cross section 1 mm2 is stretched between the support 40.05 cm apart. In its fundamental mode, it vibrates with a frequency 1000/64 Hz. The young's modulus of the wire is 10x N/m2, then find x :

Solution

## For fundamental frequency, μ=3.2 gm40 cm=3.2×10−340×10−2=3.2400=324000 kg/m l=λ2⇒λ=2l     ...(1) f=vλ=12l√Tμ ⇒100064=12×40×10−2√T32/4000 [100064×2×40×10−2]2324000=T 1000064×324000=T     ⇒T=108 N Now, y=10/810−6.05×10−240×10−2=107840(.05)=109 N/m2

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