AP EAMCET · PHYSICS · Oscillations
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
- A \(\frac{12}{25}\)
- B \(\frac{24}{25}\)
- C \(\frac{35}{24}\)
- D \(\frac{25}{12}\)
Answer & Solution
Correct Answer
(A) \(\frac{12}{25}\)
Step-by-step Solution
Detailed explanation
If both force acts simultaneously, then time period \(T=t_1 t_2\). \[ =\frac{4}{5} \times \frac{3}{5}=\frac{12}{25} \mathrm{sec} \]
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