Question:
Position of a \(2\,\text{kg}\) mass moving along the \(x\)-axis is given by
\[
x=2\cos[(2\,\text{s}^{-1})t]\ \text{m}.
\]
Then maximum kinetic energy of the mass in joule is
Position of a \(2\,\text{kg}\) mass moving along the \(x\)-axis is given by
\[
x=2\cos[(2\,\text{s}^{-1})t]\ \text{m}.
\]
Then maximum kinetic energy of the mass in joule is
Show Hint
For SHM,
\[
v_{\max}=A\omega
\]
and
\[
K_{\max}=\frac{1}{2}mA^2\omega^2.
\]
Maximum kinetic energy occurs at the mean position.
Updated On: Jun 18, 2026
- \(4\)
- \(8\)
- \(12\)
- \(16\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Compare with the standard SHM equation.
The given displacement is \[ x=2\cos(2t). \] Comparing with \[ x=A\cos(\omega t), \] we get \[ A=2\,\text{m} \] and \[ \omega=2\,\text{s}^{-1}. \]
Step 2: Find the maximum velocity.
For simple harmonic motion, \[ v_{\max}=A\omega. \] Therefore, \[ v_{\max}=2\times 2 \] \[ v_{\max}=4\,\text{m s}^{-1}. \]
Step 3: Calculate the maximum kinetic energy.
The maximum kinetic energy is \[ K_{\max}=\frac{1}{2}mv_{\max}^2. \] Given, \[ m=2\,\text{kg}. \] Hence, \[ K_{\max} = \frac{1}{2}(2)(4)^2 \] \[ K_{\max}=16\,\text{J}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{16\ \text{J}} \]
The given displacement is \[ x=2\cos(2t). \] Comparing with \[ x=A\cos(\omega t), \] we get \[ A=2\,\text{m} \] and \[ \omega=2\,\text{s}^{-1}. \]
Step 2: Find the maximum velocity.
For simple harmonic motion, \[ v_{\max}=A\omega. \] Therefore, \[ v_{\max}=2\times 2 \] \[ v_{\max}=4\,\text{m s}^{-1}. \]
Step 3: Calculate the maximum kinetic energy.
The maximum kinetic energy is \[ K_{\max}=\frac{1}{2}mv_{\max}^2. \] Given, \[ m=2\,\text{kg}. \] Hence, \[ K_{\max} = \frac{1}{2}(2)(4)^2 \] \[ K_{\max}=16\,\text{J}. \]
Step 4: Final conclusion.
Therefore, \[ \boxed{16\ \text{J}} \]
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