The speed of a transverse wave passing through a string of length 60 cm and mass 10g is 60 ms-1. The area of cross-section of the wire is 2.0 mm2 and its Young's modulus is 1.2 * 1011 Nm-2. The extension of the wire over its natural length due to its tension will be x * 10-5 m. The value of x is________.

m = 10g l=50cm A = 2mm2 Y = 1.2 * 1011N/m2 Δx=x*10−5m As, TA=YΔxl Δx=TlAY Tl=V2m = V2mAY =3600*10*10−32*10−6*1.2*1011=1800*10−3*1061.2 =1812*10−4=32*10−4=15*10−5m So, x = 15

The speed of a transverse wave passing through a string of length 60 cm and mass 10g is 60 ms^-1. The area of cross-section of the wire is 2.0 mm² and its Young's modulus is 1.2 * 10¹¹ Nm^-2. The extension of the wire over its natural length due to its tension will be x * 10^-5 m. The value of x is________.

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Payal Gupta | Contributor-Level 10

6 months ago

m = 10g

$l = 50 c m$

A = 2mm²

Y = 1.2 * 10¹¹N/m²

$Δ x = x * 1 0^{- 5} m$

As, $\frac{T}{A} = Y \frac{Δ x}{l}$

$Δ x = \frac{T l}{A Y}$

$T l = V^{2} m$

= $\frac{V^{2} m}{A Y}$

$= \frac{3600 * 10 * 1 0^{- 3}}{2 * 1 0^{- 6} * 1.2 * 1 0^{11}} = \frac{1800 * 1 0^{- 3} * 1 0^{6}}{1.2}$

$= \frac{18}{12} * 1 0^{- 4} = \frac{3}{2} * 1 0^{- 4} = 15 * 1 0^{- 5} m$

So, x = 15

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