Question

A particle executing SHM of amplitude 5cm and period 6s How long will it take to move from one end of its path on one side of mean position to a position 2.5cm on same side of mean position

Answer 1

Answer:

X = A sin w t = 5 sin w t       where w is angular frequency

w = 2 * pi * f = 2 * pi / T = 2 * pi / 6 = pi / 3 = 1.047 / s

X = 5 sin (1.047 t)

Also, we can write X = 5 cos w t          which is just using a phase angle

of pi / 2 in the first equation and means that we measure time from the

point of maximum extension instead of from the midpoint of motion

So we have 2.5 = 5 cos 1.047 t2       since we started at t1 = 0

cos 1.047 t2 = .5          

1.047 t2  = 60 deg = 1.047 rad

t2 = 1 sec   for the elapsed time