An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with frequency of $(\frac{25}{\pi}) Hz$. At the position $x = 0.04 m$, the object has kinetic energy 1 J and potential energy 0.6 J. The amplitude of oscillation is
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Total energy in SHM remains constant and is equal to potential energy at the extreme position.
Step 1: Understanding the Question:
The total energy in SHM is the sum of Kinetic Energy (KE) and Potential Energy (PE) at any instant. Total energy is also equal to the maximum potential energy at the extreme position. Step 2: Key Formula or Approach:
1. Total Energy $E = KE + PE$
2. $E = \frac{1}{2} m \omega^2 A^2$
3. Angular frequency $\omega = 2\pi f$ Step 3: Detailed Explanation:
Given:
$m = 0.2 kg$
$f = \frac{25}{\pi} Hz \implies \omega = 2\pi \cdot \frac{25}{\pi} = 50 rad/s$
At $x = 0.04 m$, $KE = 1 J$ and $PE = 0.6 J$.
Total Energy $E = 1 + 0.6 = 1.6 J$
Using the total energy formula:
\[ 1.6 = \frac{1}{2} \times 0.2 \times (50)^2 \times A^2 \]
\[ 1.6 = 0.1 \times 2500 \times A^2 \]
\[ 1.6 = 250 \times A^2 \]
\[ A^2 = \frac{1.6}{250} = \frac{16}{2500} \]
Taking the square root:
\[ A = \sqrt{\frac{16}{2500}} = \frac{4}{50} = \frac{8}{100} = 0.08 m \] Step 4: Final Answer:
The amplitude of oscillation is 0.08 m.