Verified Solution • 2025–26 Session moderate

QNEET-2017-Ph-1005: NEET 2017 Physics - Class 11 NEET-2017-Ph-1005 Solution: A particle executes linear simple harmonic motion with an…

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 equal to that of its acceleration. Then its time period in second is 5π\frac{\sqrt{5}}{\pi} 52π\frac{\sqrt{5}}{2\pi} 4π3\frac{4\pi}{\sqrt{3}} 2π3\frac{2\pi}{\sqrt{3}}

Multiple Choice Options

  • A 52π{{\sqrt 5 } \over {2\pi }} 2 � 5 _ _
  • B 4π5{{4\pi } \over {\sqrt 5 }} 5 _ 4 � _
  • C 2π3{{2\pi } \over {\sqrt 3 }} 3 _ 2 � _
  • D 5π{{\sqrt 5 } \over \pi } � 5 _ _

Step-by-Step Verified Solution

Given, Amplitude A = 3 cm When particle is at x = 2 cm According to question, magnitude of velocity = acceleration _ A 2 _ x 2 = x _ 2 ( 3 ) 2 _ ( 2 ) 2 = 2 ( 2 � T ) 5 = 4 � T _ T = 4 � 5

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