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A block of mass 2m is hanging at the lower and of a ropeof mass m and length the l, the other end being fixed to the ceililng. A pulse of wavelength lamda_(0) is produced at the lower of the rope. The time taken by the pulse to reach the other end of the rope is: |
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Answer» `2sqrt((l)/(g))(sqrt(3 )-1)` Then `v_(x) = sqrt((2M + (max)/(l))(g)/(m//l))= sqrt((2l + x))g` Then time taken to go from bottom to top will be T = `underset(0)overset(L)(INT)(DX)/(sqrt((2l +x))g) = 2sqrt((l)/(g))(sqrt(3) -sqrt(2))`. |
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