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Question

Physics Question on Oscillations

An elastic string of unstretched length LL and force constant kk is stretched by a small length xx. It is further stretched by another small length yy. The work done in the second stretching is

A

1/2Ky21/2 Ky^2

B

1/2Ky(2x+y)1/2 Ky(2x + y)

C

1/2K(x2+y2)1/2 K(x^2 + y^2)

D

1/2K(x+y)21/2 K(x +y)^2

Answer

1/2Ky(2x+y)1/2 Ky(2x + y)

Explanation

Solution

In the string elastic force is conservative in nature. W=ΔU\therefore W =-\Delta U Work done by elastic force of string. W=(UFUi)=UiUFW =-\left( U _{ F }- U _{ i }\right)= U _{ i }- U _{ F } W=12kx2k2(x+y)2W =\frac{1}{2} kx ^{2}-\frac{ k }{2}( x + y )^{2} =12kx212k(x2+y2+2xy)=\frac{1}{2} kx ^{2}-\frac{1}{2} k \left( x ^{2}+ y ^{2}+2 xy \right) =12kx212ky212kx212k(2xy)=\frac{1}{2} kx ^{2}-\frac{1}{2} ky ^{2}-\frac{1}{2} kx ^{2}-\frac{1}{2} k (2 xy ) =kxy12ky2=- kxy -\frac{1}{2} ky ^{2} Therefore, the work done against elastic force Wexternal =W=ky2(2x+y)W _{\text {external }}=- W =\frac{ ky }{2}(2 x + y ).