A long straight and massless rod pivots about one end in a vertical plane. In configuration I, two small identical masses are attached to the free end, in the second configuration one mass is attached at the centre and the other is attached to the free end. What is the ratio of frequency of small oscillations of configuration / to that of configuration II?Correct answer is '1.09'. Can you explain this answer?

Question

Answer

Introduction:
In this problem, we are given two different configurations of a long, straight, and massless rod pivoting about one end in a vertical plane. We need to find the ratio of the frequency of small oscillations in configuration I to that in configuration II.

Configuration I:
In configuration I, two small identical masses are attached to the free end of the rod. Let's denote the masses as m1 and m2.

Configuration II:
In configuration II, one mass (m1) is attached at the center of the rod, and the other mass (m2) is attached to the free end.

Derivation:
To find the ratio of the frequencies, we need to derive the expression for the frequency of oscillation in each configuration.

Configuration I:
In configuration I, the masses m1 and m2 are attached to the free end. Let's assume the distance between the pivot point and m1 is x1, and the distance between the pivot point and m2 is x2.

The moment of inertia of the system about the pivot point is given by:
I1 = m1 * x1^2 + m2 * x2^2

The torque acting on the system is due to the gravitational force acting on the masses. The torque equation is given by:
τ1 = (m1 * g * x1) + (m2 * g * x2)

Using the equation τ = I * α, where α is the angular acceleration, we can write:
(m1 * g * x1) + (m2 * g * x2) = (m1 * x1^2 + m2 * x2^2) * α

The angular acceleration α can be written as α = -ω^2 * θ, where ω is the angular velocity and θ is the angular displacement.

Substituting the value of α, we get:
(m1 * g * x1) + (m2 * g * x2) = (m1 * x1^2 + m2 * x2^2) * (-ω^2 * θ)

The equation can be rearranged as:
(m1 * g * x1) + (m2 * g * x2) = -(m1 * x1^2 + m2 * x2^2) * ω^2 * θ

Comparing this equation with the equation of simple harmonic motion, we can find the expression for the frequency of oscillation in configuration I:
ω1 = √((g * (m1 * x1 + m2 * x2)) / (m1 * x1^2 + m2 * x2^2))

Configuration II:
In configuration II, the mass m1 is attached at the center of the rod, and the mass m2 is attached to the free end. Let's assume the distance between the pivot point and the center of the rod is x, and the distance between the center of the rod and m2 is y.

The moment of inertia of the system about the pivot point is given by:
I2 = (1/3) * m1 * x^2 + m2 * y^2

The torque acting on the system is due to the gravitational force acting on the masses. The torque equation is given by:

Similar Physics Doubts

Two masses m1 and m2 are suspended together by a massless spring of spring constant k. When the masses are in equilibrium m1 is removed without disturbing the system.

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