Question:
A particle undergoing simple harmonic motion has an amplitude of \(10\,\text{cm}\). When the particle is at a displacement of \(6\,\text{cm}\) from the center, then the ratio of its kinetic energy to potential energy is
A particle undergoing simple harmonic motion has an amplitude of \(10\,\text{cm}\). When the particle is at a displacement of \(6\,\text{cm}\) from the center, then the ratio of its kinetic energy to potential energy is
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In SHM,
\[
K\propto (A^2-x^2)
\]
and
\[
U\propto x^2
\]
So the ratio \(K:U\) can be found directly without calculating constants.
Updated On: Jun 26, 2026
- \(3:2\)
- \(9:4\)
- \(16:9\)
- \(4:3\)
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The Correct Option is C
Solution and Explanation
Step 1: Write the expressions for total energy and potential energy in SHM.
For a particle executing SHM, \[ \text{Total Energy}=\frac{1}{2}kA^2 \] Potential energy at displacement \(x\) is \[ U=\frac{1}{2}kx^2 \] Hence kinetic energy is \[ K=\frac{1}{2}k(A^2-x^2) \]
Step 2: Substitute the given values.
Given, \[ A=10\,\text{cm} \] and \[ x=6\,\text{cm} \] Therefore, \[ K\propto A^2-x^2 \] \[ =10^2-6^2 \] \[ =100-36 \] \[ =64 \] Also, \[ U\propto x^2 \] \[ =6^2 \] \[ =36 \]
Step 3: Find the ratio \(K:U\).
\[ K:U=64:36 \] Dividing by \(4\), \[ K:U=16:9 \]
Step 4: Final conclusion.
Hence, the required ratio is \[ \boxed{16:9} \]
For a particle executing SHM, \[ \text{Total Energy}=\frac{1}{2}kA^2 \] Potential energy at displacement \(x\) is \[ U=\frac{1}{2}kx^2 \] Hence kinetic energy is \[ K=\frac{1}{2}k(A^2-x^2) \]
Step 2: Substitute the given values.
Given, \[ A=10\,\text{cm} \] and \[ x=6\,\text{cm} \] Therefore, \[ K\propto A^2-x^2 \] \[ =10^2-6^2 \] \[ =100-36 \] \[ =64 \] Also, \[ U\propto x^2 \] \[ =6^2 \] \[ =36 \]
Step 3: Find the ratio \(K:U\).
\[ K:U=64:36 \] Dividing by \(4\), \[ K:U=16:9 \]
Step 4: Final conclusion.
Hence, the required ratio is \[ \boxed{16:9} \]
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