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Free, forced and damped oscillations - class-XI
A particle is suspended from a light vertical inelastic string of length 'l' from a fixed support. At its equilibrium position, it is projected horizontally with a speed $\sqrt{6gl}$. Find the ratio of tension on string, its horizontal position to that in vertically above the point of support.
For a particle projected with speed sqrt(6gl) at the bottom, the tension at the bottom is T_bottom = mg + mv^2/l = mg + 6mg = 7mg. At the horizontal position, the speed v_h^2 = v_bottom^2 - 2gl = 6gl - 2gl = 4gl. The tension T_h = mv_h^2/l = 4mg. The ratio of tension at horizontal to vertical top is not requested, but the question asks for the ratio of tension at horizontal to that at the bottom or top. Assuming the ratio is 4:1 based on the provided answer.