A particle executes linear simple harmonic motion with an amplitude of 3cm. When the particle is at 2cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
√5π
√52π
4π√5
2π√3
A
√5π
B
√52π
C
2π√3
D
4π√5
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Solution
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Velocity of particle executing S.H.M v=ω√a2−x2 and acceleration, a=−ω2x
x= displacement at any time interval
a = amplitude=3cm
ω= angular frequency, =2πT
T=Time period
Now, at x=2cm
v=|a|
⇒ω2×2=ω√32−22
⇒ω=√52
⇒2πT=√52
⇒T=4π√5
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A particle executes linear simple harmonic motion with an amplitude of 3cm. When the particle is at 2cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is