SHMHard
Question
Two particle A and B perform SHM along the same the same straight line the the same amplitude ′a′ same frequency ′f′ and same equilibrium position ′O′. The greatest distance between them is found to be 3a/2. At some instant of time they have same displacement from mean position. What is this displacement ?
Options
A.a / 2
B.a √7 / 4
C.√3 / a2
D.3a / 4
Solution
z1 = a sin , x2 = a sin(ωt + φ)
Greatest distance
= |x2 - x1|max = 2a sin
Now according to question x1 = x2
⇒ a sin ωt = a sin (ωt + φ)
⇒ π - ωt = ωt + φ ⇒ ωt =
⇒ x1 = a sin
= a cos 
Greatest distance
= |x2 - x1|max = 2a sin
Now according to question x1 = x2
⇒ a sin ωt = a sin (ωt + φ)
⇒ π - ωt = ωt + φ ⇒ ωt =
⇒ x1 = a sin
= a cos Create a free account to view solution
View Solution FreeMore SHM Questions
If a simple pendulum has significant amplitude(up to a factor of 1/e of original) only in theperiod between t = 0 s to t...Displacement of a particle is x = 3 sin2t + 4cos 2t , the amplitude and the max. velocity will be :...A particle of mass m executing SHM makes f oscillation per second. The difference of its kinetic energy when at the cent...The equation of a simple harmonic wave is given byy = 3 sin (50t - x) where x and y are in metres and t is in seconds. T...Acceleration a versus time t graph of a body in SHM is given by a curve shown below. T is the time...