A particle executes SHM, with an amplitude of 10cm. When the particle is at 6cm from the mean position, the magnitude of its velocity is equal to twice of its acceleration. Find the time period of SHM(in seconds)
A
4.1s
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B
9.42s
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C
1.17s
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D
14.8s
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Solution
The correct option is B9.42s Given,
Amplitude (A)=10cm
Position of particle from mean position (x)=6cm
Since, particle executes SHM, we can use v=ω√A2−x2 ... (1) a=ω2x ... (2) ∴ from (1) and (2) in |v|=2|a| (Given in the question)
we get ω√A2−x2=2ω2x ⇒ω=√A2−x24x2 ∴ Time period (T)=2πω =2π√4x2A2−x2=2×3.14×2×6√(10)2−(6)2 =2×3.14×2×68 =9.42s