TS EAMCET · Physics · Oscillations
The amplitude of a particle executing simple harmonic motion is 6 cm. The distance of the point from the mean position at which the ratio of the potential and kinetic energies of the particle becomes 4:5 is
- A 6 cm
- B 4 cm
- C 3 cm
- D 2 cm
Answer & Solution
Correct Answer
(B) 4 cm
Step-by-step Solution
Detailed explanation
\(\frac{PE}{KE} = \frac{\frac{1}{2} k x^2}{\frac{1}{2} k (A^2 - x^2)} = \frac{x^2}{A^2 - x^2}\) \(\frac{x^2}{A^2 - x^2} = \frac{4}{5}\) \(5x^2 = 4(A^2 - x^2)\) \(9x^2 = 4A^2\) \(x = \frac{2}{3} A\) \(x = \frac{2}{3} \times 6 \text{ cm}\) \(x = 4 \text{ cm}\)
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