A and B are two points on the path of a particle executing SHM in a straight line, at which its velocity is zero. Starting from a certain point X `(AXltBX)` the particle crosses this pint again at successive intervals of 2 seconds and 4 seconds with a speed of `5m//s` Q. Amplitude of oscillation is
A
`10sqrt(3)m`
B
`(10)/(sqrt(3)pi)m`
C
`(10sqrt(3))/(pi)m`
D
can't determined
Text Solution
Verified by Experts
The correct Answer is:
C
Maximum speed `= Aomega = ((10sqrt(3))/(pi)) ((2pi)/(6)) = (10)/(sqrt(3)) m//s`
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