A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω . The amplitude of oscillation is gra - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time:

Solution & Step-by-step Explanation

The coin leaves the platform when the downward acceleration of the platform exceeds the acceleration due to gravity g .This occurs at the highest point of the oscillation where the downward acceleration is maximum: a_{ma x} = ω^{2} A .The condition for leaving contact is: ω^{2} A \geq g ⟹ A \geq \frac{g}{ω ^{2}} Thus, it happens for the first time at the highest position when the amplitude reaches g / ω^{2} .

Try it yourself before checking the explanation above.

A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency ω . The amplitude of oscillation is gradually increased. The coin will leave contact with the platform for the first time:

at the highest position of the platform

at the mean position of the platform

for an amplitude of g / ω^{2}

for an amplitude of g /2 ω^{2}

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