1) There are three simple pendulums, where the length of the massless strings, size of the masses and angles with respect to equilibrium positions are identical and frictional forces are negligible.

In a, the mass of the ball is 3kg. In b, the mass of the ball is 4kg. In c, the mass of the ball is 6kg.

i) Increasingly rank the period of these pendulums ii) Increasingly rank the tension of these pendulums iii) Increasingly rank the angular frequency of these pendulums

2) If the length of massless string in B is halved and the length of massless string in C is quadrupled,

i) Increasingly rank the period of these pendulums ii) Increasingly rank the tension of these pendulums iii) Increasingly rank the angular frequency of these pendulums

3) If pendulum a experiences drag force with the some proportionality constant b, when will it experience the most drag force.

Answers:

1) i) Period (T)= 2pi /[ (g/L)^(1/2) ] the weight of pendulums do not affection the period. So Ta=Tb=Tc

ii) Tension=mgcosx . The greater the mass, the greater the tension is.Tension a<Tension b<Tension ciii) angular frequency= (g/L)^(1/2) because the length of pendulums are the same, angular frequency a= angular frequency b= angular frequency c 2) i) Period (T)= 2pi /[ (g/L)^(1/2) ]because period is directly proportional to length. Tb<Ta<Tc

ii) As Tension=mgcosx, it does not depend on the length of string. So, the rank of tension remains unchanged Tension a<Tension b<Tension c. iii) Angular frequency= (g/L)^(1/2) because angular frequency is inversely proportional to length of string, angular frequency c <angular frequency a< angular frequency b 3) The magnitude of frictional force is proportional to the speed of the object. Pendulum a is the fastest when it is perpendicular to Fg therefore it experiences the greatest frictional force then.