Who Wants To Draw A Diagram?
A metre rule is clamped to the top of a bench so that most of its length overhangs and is free to vibrate in a vertical plane with simple harmonic motion. It oscillates with a frequency of Hz. A small mass rests on the end of the metre rule and oscillates with it. The rule is oscillated with different amplitudes. As the amplitude is increased, it is observed that at one point in the cycle of motion of the rule and the mass, the mass loses contact with the rule. Calculate, in metres, the minimum amplitude of oscillation at which the loss of contact occurs.
The answer is 0.0099.
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1 solution
From hookes law, F=kx=m(w^2)x=m×4×(3.1416)^2×(f^2)x
now, the ball will stay on the ruler until the ruler vibrates in such an amplitude that causes the restoring force to go beyond the ball's weight. So the minimum amplitude to disconnect the ball will be just beyond the amplitude of force equal to the weight of the ball.
Therefore,mg=m×4×(3.1416)^2×(f^2)x
now we cancel m from both sides and plug in f=5. Solving for x yields x=0.0099 So, our required amplitude is just beyond this magnitude and that is x=0.01 metre