Question

A square object of mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner (Fig. 2). If it is rotated slightly to the left and then released, at what frequency will it swing back and forth? ( 20 points)

Answer 1

The frequency at which the square object will swing back and forth can be determined using the formula for the period of a simple pendulum. The period depends on the length of the pendulum and the acceleration due to gravity. In this case, the length of the pendulum is equal to the diagonal of the square, which is L√2.

Step-by-step explanation:

The frequency at which the square object will swing back and forth can be determined using the formula for the period of a simple pendulum. In this case, the length of the pendulum is equal to the diagonal of the square, which is L√2.

The period of a simple pendulum is given by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

Substituting the length of the pendulum (L√2) into the formula, we get:
T = 2π√((L√2)/g) = 2π√2(L/g)

Therefore, the frequency of the square object's swing back and forth is the reciprocal of the period, which is f = 1/T = 1/(2π√2(L/g)) = √2/(4π) × √(g/L)