A body executes simple harmonic motion under the action of a force $F_1$ with a time period $\frac{4}{5}…

A body executes simple harmonic motion under the action of a force $F_1$ with a time period $\frac{4}{5} \mathrm{~s}$. If the force is changed to $F_2$ it executes SHM with time period $\frac{3}{5} \mathrm{~s}$. If both the forces $F_1$ and $F_2$ act simultaneously in the same direction on the body, its time period in seconds is
  1. $\frac{12}{25}$
  2. $\frac{24}{25}$
  3. $\frac{35}{24}$
  4. $\frac{25}{12}$

Solution

If both force acts simultaneously, then time period $T=t_1 t_2$. $ =\frac{4}{5} \times \frac{3}{5}=\frac{12}{25} \mathrm{sec} $

Asked in: AP EAMCET 2002

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