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
(a)`10sqrt(3)m`
B
(b)`(10)/(sqrt(3)pi)m`
C
(c)`(10sqrt(3))/(pi)m`
D
can't determined
Text Solution
Verified by Experts
The correct Answer is:
C
The given diagram can be represented in circular motion In shoqwn diagram `theta = ((2pi)/(6)) (2) = (2pi)/(3)` Let `OX = x_(0)` and amplitude `= OP= A` Now `(x_(0))/(A) = cos"(theta)/(2)` so `x_(0) = (A)/(2)` As `v = omegasqrt(A^(2) - X_(2)^(2))` so `5 = (2pi)/(6)sqrt(A^(2) - (A^(2))/(4)) rArr A = (10sqrt(3))/(pi) m`
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