A particle executes I inear simple harmonic motion with amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is In

The correct answer is Option 1. In simple harmonic motion (SHM), the magnitude of velocity is given by v = ω√(A² - x²) and the magnitude of acceleration is a = ω²x, where A is the amplitude, x is the displacement, and ω is the angular frequency.

According to the question:

• Amplitude (A) = 2 cm
• Displacement (x) = 1 cm
• Condition: |v| = |a|

Substituting the values into the condition:
ω√(2² - 1²) = ω²(1)
√(3) = ω

Since the angular frequency ω = 2π/T, we equate:
2π/T = √3
T = 2π/√3

The question asks for the time period in seconds. As per the provided correct option, the value corresponds to 2π/√3. The other options are marked as "None," making Option 1 the only valid numerical choice derived from the SHM equations.