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A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is numerically equal to that of its acceleration. Then its time period in seconds is

Asked in NEET 2017 · v-x and a-x relations

• (1) (√5)/(2π)
• (2) (4π)/(√5)
• (3) (2π)/(√3)
• (4) (√5)/π

Step-by-step solution

Why the other options are wrong

• (1) comes from is the relation T=2π/ω turned upside down; with ω=√5/2≈ 1.1 rad s⁻¹ the period is about 5.6 s, not the 0.36 s this gives
• (3) comes from needs ω=√3 rad s⁻¹; the condition ω√A²-x²=ω²x fixes ω=√5/2 whichever way it is rearranged, so √3 cannot appear
• (4) comes from is the same inversion doubled, giving about 0.71 s; it is still far below 2π/ω≈ 5.6 s

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