Simple Harmonic Motion
Classical Mechanics
Level
4
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<PB) at successive intervals of 0.5 and 1.5 secs with a speed of 3 root(2) mps.Determine the maximum speed in mps.
The answer is 6.
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1 solution
first write the equation of a particle executing SHM , y = Asin(wt) ...now on differentiating we get the velocity at time " t " which is v= wAcoswt .... here we are asked to find the max velocity and max velocity = wA .. (it is possible at its mean position ) .... v= max velo * cos( wt) ... now w=2pi/T where T is the time period ... here time period is 2 * (1.5 + 0.5) = 4 and cos wt = 1/root(2) ... 3 * root(2) * root(2) = max velocity = 6mps ....