Question:
A block is executing simple harmonic motion and its position is given by \(x = A \cos(\omega t + \phi)\). At time \(t = 0\), the ratio of the kinetic energy to the potential energy is \(\frac{1}{3}\). The magnitude of the phase angle \(\phi\) is:
A block is executing simple harmonic motion and its position is given by \(x = A \cos(\omega t + \phi)\). At time \(t = 0\), the ratio of the kinetic energy to the potential energy is \(\frac{1}{3}\). The magnitude of the phase angle \(\phi\) is:
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In SHM, \(\frac{KE}{PE} = \tan^2 \phi\) at any instant. Use position at t=0 to find phase angle if KE/PE ratio is known.
Updated On: Jun 19, 2026
- 60°
- 30°
- \(\tan^{-1}\left(\frac{2}{3}\right)\)
- \(\tan^{-1}\left(\frac{3}{2}\right)\)
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The Correct Option is B
Solution and Explanation
Step 1: Kinetic and potential energy in SHM.
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2),\quad PE = \frac{1}{2} m \omega^2 x^2 \] At \(t=0, x = A \cos\phi\). The ratio given is: \[ \frac{KE}{PE} = \frac{A^2 - A^2 \cos^2 \phi}{A^2 \cos^2 \phi} = \frac{\sin^2 \phi}{\cos^2 \phi} = \tan^2 \phi \]
Step 2: Solve for \(\phi\).
\[ \tan^2 \phi = \frac{KE}{PE} = \frac{1}{3} \Rightarrow \tan \phi = \frac{1}{\sqrt{3}} \Rightarrow \phi = 30^\circ \]
Step 3: Conclusion.
The phase angle is 30°.
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2),\quad PE = \frac{1}{2} m \omega^2 x^2 \] At \(t=0, x = A \cos\phi\). The ratio given is: \[ \frac{KE}{PE} = \frac{A^2 - A^2 \cos^2 \phi}{A^2 \cos^2 \phi} = \frac{\sin^2 \phi}{\cos^2 \phi} = \tan^2 \phi \]
Step 2: Solve for \(\phi\).
\[ \tan^2 \phi = \frac{KE}{PE} = \frac{1}{3} \Rightarrow \tan \phi = \frac{1}{\sqrt{3}} \Rightarrow \phi = 30^\circ \]
Step 3: Conclusion.
The phase angle is 30°.
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