The angular velocities of three bodies in simple harmonic motion are \\omega_{1}, \\omega_{2}, \\omega_{3} with their respective amplitudes as A_{1}, A_{2}, A_{3}. If all the three bodies have same mass and velocity, then

Correct Answer is : A1ω1=A2ω2=A3ω3

A_{1}^{2} \\omega_{1}^{2}=A_{2}^{2} \\omega_{2}^{2}=A_{3}^{2} \\omega^{2}

A_{1}^{2} \\omega_{1}=A_{2}^{2} \\omega_{2}=A_{3}^{2} \\omega_{3}

A_{1} \\omega_{1}^{2}=A_{2} \\omega_{2}^{2}, A_{3} \\omega_{3}^{2}

A_{1} \\omega_{1}=A_{2} \\omega_{2}=A_{3} \\omega_{3}

Manipal Entrance Test 2013 Physics Paper

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The angular velocities of three bodies in simple harmonic motion are ω1,ω2,ω3 with their respective amplitudes as A1,A2,A3. If all the three bodies have same mass and velocity, then

A12ω12=A22ω22=A32ω2
A12ω1=A22ω2=A32ω3
A1ω12=A2ω22,A3ω32
A1ω1=A2ω2=A3ω3

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