A bob attached to a string of length L = 1.25 m, initially found at the equilibrium position, is given an initial velocity v = 0.8 m/s. The maximum displacement angle, Omax, is: (Take g=10 m/s)

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Answer 1

The maximum displacement angle is given as  13°

L = length = 1.25 meters

Initial velocity = 0.8m/s

displacement = ∅max

Work done is given as change in kinetic energy

[tex]\frac{1}{2} mvi^2+0=0+mgh\\\\h = \frac{vi^2}{2g} \\\\h = \frac{0.8^2}{2*10} =\frac{0.64}{20} =0.032 meteres[/tex]

L-h

= 1.25-0.032

= 1.218 meters

Cos∅max = x/L

[tex]=\frac{1.218}{1.25} \\\\= 0.9744\\\\[/tex]

∅max = Cos⁻¹(0.9744)

= 12.99°

≈ 13°

Therefore the maximum displacement angle is given as  13°

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