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
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`
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. The ratio (AX)/(BX) is
X_(1) and X_(2) 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(X_(1)XltX_(2)X) then particle crosses this point again at successive intervals of 2s and 4s with a speed of 5m//s . The time period of SHM is
A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P(APltBP) at successive intervals of 0.5s and 1.5 s with a speed of 3m/s.
A particle is executing SHM in straight line starting its motion at t_(1)=0 with zero initial phase.It crosses a point during the motion at successive intervals at t_(2)=t and t_(3)=2t with a speed v. If its acceleration amplitude is (2 pi V)/(Nt) then N is equal to
If a particle executing S.H.M. with amplitude A and maximum velocity vo, then its speed at displacement A/2 is
A particle executes SHM in a straight line. In the first second starting from rest it travels distance a and in the next second it travels distance a and in the next second it travels distance d in the same direction. The amplitude of SHM is :
(i) The acceleration (a) of a particle moving along a straight line is related to time (t) as per the differential equation (d^(2)a)/(dt^(2))=-ba. b is a positive constant. Is the particle performing SHM? If yes, what is the time period? (ii) A particle is executing SHM on a straight line. A and B are two points at which its velocity is zero. It passes through a certain point P (AP gt PB) with a speed of 3 m//s at times recorded as t = 0, 0.5 s, 2.0 s, 2.5 s, 4.0 s, 4.5 s......Determine the maximum speed of the particle and also the ratio AP//PB .
A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))
A particle moving with a uniform acceleration along a straight line covers distances a and b in successive intervals of p and q second. The acceleration of the particle is
