Answer:
Let A be the displacement amplitude and co be the angular frequency of S.H.M. Then Maximum velocity, \[{{\upsilon }_{0}}=\omega A\]\[\therefore \] \[\omega ={{\upsilon }_{0}}/A\]Maximum acceleration, \[{{a}_{0}}={{\omega }^{2}}A={{\left( \frac{{{\upsilon }_{0}}}{A} \right)}^{2}}A=\frac{\upsilon _{0}^{2}}{A}\] \[\therefore \] Displacement amplitude, \[A=\frac{\upsilon _{0}^{2}}{{{a}_{0}}}.\]
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