MHT CET · Physics · Oscillations
The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in mean position. If amplitude of motion is 25 cm and the mass of the particle is 5.12 kg , the period of oscillation is
- A \(\frac{\pi}{5} \mathrm{~s}\)
- B \(2 \pi \mathrm{~s}\)
- C \(20 \pi \mathrm{~s}\)
- D \(5 \pi \mathrm{~s}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{5} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Given: K.E. \(=16 \mathrm{~J}, \mathrm{~A}=25 \mathrm{~cm}=0.25 \mathrm{~m}\), \(\mathrm{m}=5.12 \mathrm{~kg}\)
The kinetic energy at mean position is equal to the total energy of the particle.
\(\begin{array}{ll}
\therefore & \text { Total energy, } E=\frac{1}{2} \mathrm{~m} \omega \\
\therefore & \omega^2=\frac{16 \times 2}{5.12 \times(0.25)^2}=10^2 \\
\therefore & \omega=10 \\
& \mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{10}=\frac{\pi}{5} \mathrm{~s}
\end{array}\)
The kinetic energy at mean position is equal to the total energy of the particle.
\(\begin{array}{ll}
\therefore & \text { Total energy, } E=\frac{1}{2} \mathrm{~m} \omega \\
\therefore & \omega^2=\frac{16 \times 2}{5.12 \times(0.25)^2}=10^2 \\
\therefore & \omega=10 \\
& \mathrm{~T}=\frac{2 \pi}{\omega}=\frac{2 \pi}{10}=\frac{\pi}{5} \mathrm{~s}
\end{array}\)
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