Q156 / 200 · AP EAMCET 2002 · Physics

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

  • Correct answer
    A
    $\frac{12}{25}$
  • B
    $\frac{24}{25}$
  • C
    $\frac{35}{24}$
  • D
    $\frac{25}{12}$

Solution

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

Hence, the correct answer is option: 'A'.

Relevant Questions