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 = 0.17 kg) and the other of metal (mass = 0.85). They are set into oscillation and execute simple harmonic motion. Determine the period of (a) the wood pendulum and (b) the metal pendulum.
The answer is 1.64 s for both. Can someone show the steps to see how to get this?
To find the periods of the wood and metal pendulums, we can use the formula for the period of a physical pendulum:
T = 2π * √(I / (m * g * d))
where:
T is the period of the pendulum,
I is the moment of inertia of the pendulum about its axis of rotation,
m is the mass of the pendulum,
g is the acceleration due to gravity, and
d is the distance from the axis of rotation to the center of mass.
Given that both pendulums are made from meter sticks and are identical in all respects except for the mass, we can assume they have the same length and the same distance between the axis of rotation and the center of mass.
For a uniform meter stick suspended from one end, the moment of inertia about the axis of rotation is given by:
I = (1/3) * m * L^2
where L is the length of the meter stick.
Let's calculate the period for the wood pendulum first:
For the wood pendulum:
m = 0.17 kg
g = 9.8 m/s^2 (acceleration due to gravity)
L = 1 m (length of the meter stick)
Substituting these values into the formula, we have:
T = 2π * √((1/3) * m * L^2 / (m * g * d))
Since the mass cancels out:
T = 2π * √((1/3) * L^2 / (g * d))
For the metal pendulum:
m = 0.85 kg
Substituting the values into the formula, we have:
T = 2π * √((1/3) * L^2 / (g * d))
Since the length and distance between the axis of rotation and center of mass are the same for both pendulums, the periods will be equal.
Therefore, the period for both the wood and metal pendulum is 1.64 seconds (approximately).