Two physical pendulums (not simple pendulums) are made from meter sticks that are suspended from the ceiling at one end. The sticks are uniform and are identical in all respects, except that one is made of wood (mass = 0.16 kg) and the other of metal (mass = 0.88 kg). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.

(A) T = ? s

(B) T = ? s

Answer: A and B is 1.64 seconds.

The period is the same for the wood and metal pendulums because the mass has been eliminated mathematically in the expression for T.

Explanation: The period T is given by

T = 2π/ω, where ω is the angular frequency.

ω, angular frequency, is represented by

Sqaureroot of (m*g*L/I), where m is mass, L is the distance from pivot to the center of mass and I is the moment of inertia about the pivot.

Therefore

T = 2π / sqaureroot (m*g*L/I),

Further expressed as

T = 2π*sqaureroot (I/m*g*L),

Inertia Is given by

I=1/3mL^2 where L=length of the stick. Since the meter stick is uniform, the distance L from one end to the center of mass is Upper L = one-half Upper L 0 period.

Thus the period of oscillation is expressed as;

T=2π*sqrt (I/mg (1/2L)

=2π*sqrt (1/3m*L^2/mg (1/2L)

= 2π*sqrt (2L/3g)

=2π*sqrt (2*1/3*9.8)

= 1.64 seconds