Question

# For the damped oscillator shown in Fig, the mass of the block is $$200\ g, k = 80\ N\ m^{-1}$$ and the damping constant $$b$$ is $$40\ g\ s^{-1}$$ Calculate.(a) The period of oscillation,(b) Time period for its amplitude of vibrations to drop to half of its initial value(c) The time for the mechanical energy to drop to half initial value.

Solution

## For damped oscillation,$$(a)\quad T=\dfrac { 2\pi }{ \omega } \\ T=\dfrac { 2\pi }{ \sqrt { \dfrac { K }{ m } -{ \left( \dfrac { l }{ 2m } \right) }^{ 2 } } } \\ T=\dfrac { 2\times 3.14 }{ \sqrt { \dfrac { 80\times 1000 }{ 200 } -{ \left( \dfrac { 40\times 1000 }{ 1000\times 2\times 200 } \right) }^{ 2 } } } \\ T=\dfrac { 6.28 }{ \sqrt { 400-\dfrac { 1 }{ 100 } } } \\ T=0.314s$$$$(b)\quad F\times { A }^{ 2 }$$If amplitude gets reduced to half the force will reduce to one fourth.$$F\propto a\\ a\propto { \omega }^{ 2 }$$So $$\omega$$ will become half.$$T=\dfrac { 2\pi }{ \omega } \\ T=2\times 0.314=0.628s$$Physics

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