A particle executing SHM has maximum acceleration of 2πm/s2 and maximum velocity of 6m/s. Find its time period
A
π22s
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B
2π23s
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C
6s
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D
π3s
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Solution
The correct option is C6s Given,
Maximum acceleration (amax)=2πm/s2
Maximum velocity (vmax)=6m/s
Since, the particle executes SHM, we can use amax=ω2A ... (1) vmax=ωA ... (2)
From (1) and (2) ω=amaxvmax ⇒T=2πvmaxamax=2π×62π=6s