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. Find the time period in seconds.
A
√52π
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B
4π√5
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C
2π√3
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D
√5π
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Solution
The correct option is B4π√5 Velocity-displacement relationship is given by v=ω√A2−x2 Acceleration-displacement relationship is given by |a|=ω2x Given that, |→v|=|→a| ⇒ω√A2−x2=ω2x ⇒ω2(A2−x2)=ω4x2 Given that amplitude A=3 cm & required distance x=2cm ⇒ω2=A2−x2x2=9−44=54 ⇒ω=√52 ⇒T=4π√5 seconds (∵T=2πω)