Question

A particle executes simple harmonic motion, such that at a given time, it is at ?A/3 moving in towards equilibrium. 0.7seconds later, it is at x=0.9A moving towards equilibrium. Find the angular frequency of the particle, if it passes through equilibrium once between the two occurrences.

Repeat the above, with the particle passing through equilibrium 5times between the two occurrences.

Answer 1

Solutioin:

x = Acos?t

assume up is the positive direction

(-1/3)A = Acos?t
-1/3 = cos?t
?t = ± 1.9106 radians

As we are told that the position is negative but the velocity is positive, we can assume that the clockwise frequency is in the third quadrant. Therefore we choose the negative value

?t = -1.9106 rad

0.9A = Acos(?(t + 0.7))
0.9 = cos(?(t + 0.7))
?(t + 0.7) = ± 0.45103

As the position is positive but the velocity negative we choose the positive (first quadrant) value

?(t + 0.7) = 0.45103
?t + 0.7? = 0.45103

-1.9106 + 0.7? = 0.45103
0.7? = 2.36163
? = 3.374 rad/s.

The oscillation is given by
y = A cos(3.374 t).

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