Circular Oscillation

Classical Mechanics Level pending

Two identical balls A A and B B , each of mass 0.1 kg , 0.1\text{ kg}, are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle, as shown in the figure. The pipe is fixed in a horizontal plane.

The centers of the balls can move in a circle of radius 0.06 m . 0.06\text{ m}. Each spring has a natural length of 0.06 π m 0.06\pi\text{ m} and a force constant of 0.1 N/m . 0.1\text{ N/m}. Initially, both the balls are displaced by an angle of θ = π / 6 \theta = π/6 with respect to diameter P Q PQ of the circle, and then they are released from rest. The speed ( ( in m/s ) \text{m/s}) of ball A A when both the balls are at the two ends of the diameter P Q PQ is v A v_A .

Find ⌊ 100 v A ⌋ . \big\lfloor 100v_A \big\rfloor.

Bonus: Find the frequency of oscillation of ball B . B.


The answer is 6.

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1 solution

Let y y be the initial angular displacement when at rest.

Total Energy of the system is

E = 1 2 k 1 y 2 + 1 2 k 2 y 2 = 1 2 ( 2 k y 2 ) = k y 2 E = \dfrac 12 k_1 y^2 + \dfrac 12 k_2 y^2 = \dfrac 12( 2ky^2) = ky^2

Since, k 1 = k 2 k_1 = k_2 and y 1 = y 2 y_1 = y_2

Now, y = y 1 + y 2 = R ( θ 1 + θ 2 ) = 2 × 0.06 × ( π / 6 ) y = y_1 + y_2 = R(\theta _1 + \theta _2) = 2\times 0.06 \times (π/6)

E = 0.1 × ( 0.02 π ) 2 = 4 π 2 × 1 0 − 5 J E= 0.1 \times (0.02π)^2 = 4π^2 \times 10^{-5} J

Now, 1 5 m A v A 2 + 1 2 m B v B 2 = 4 π 2 × 1 0 − 5 \dfrac 15 m_A v_A^2 + \dfrac 12 m_B v_B^2 = 4π^2 \times 10^{-5}

m 1 = m 2 = 0.1 k g m_1 = m_2 = 0.1 kg and v A = v B v_A = v_B

0.1 v A 2 = 4 π 2 × 1 0 − 5 ⇒ v A = 2 π × 1 0 − 2 m / s 0.1 v_A^2 = 4π^2 \times 10^{-5} \Rightarrow v_A = 2π\times 10^{-2} m/s

Therefore ⌊ 100 v A ⌋ = ⌊ 6.28.. ⌋ = 6 \lfloor 100v_A \rfloor = \lfloor 6.28.. \rfloor = \boxed{6}

For frequency consider reduced mass μ \mu of the system

μ = m A m B m A + m B = m 2 = 0.05 k g \mu = \dfrac {m_A m_B }{m_A + m_B} = \dfrac m2 = 0.05 kg

k e f f = k 1 + k 2 = 0.1 + 0.1 = 0.2 N / m k_{eff} = k_1 + k_2 = 0.1 + 0.1 = 0.2 N/m

Frequency f = 1 2 π k e f f μ = 1 π H z f = \dfrac {1}{2π} \sqrt {\dfrac {k_{eff}}{\mu }} = \boxed{\dfrac {1}{π} Hz}

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