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. 60°
• B. 30°
• C. \(\tan^{-1}\left(\frac{2}{3}\right)\)
• D. \(\tan^{-1}\left(\frac{3}{2}\right)\)

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Energy expressions in SHM.
KE = ½ m ω² (A² - x²), PE = ½ m ω² x². At t=0, x = A cos φ.

Step 2: Forming the ratio.
KE/PE = (A² - A² cos²φ)/(A² cos²φ) = sin²φ/cos²φ = tan²φ.

Step 3: Solving for φ.
tan²φ = 1/3 → tan φ = 1/√3 → φ = 30°.

Step 4: Conclusion.
The phase angle is 30°.