A04. A block rests on a spring and oscillates with a frequency of 4.00 Hz and an

Question

A04. A block rests on a spring and oscillates with a frequency of 4.00 Hz and an amplitude of 7.00 cm. A tiny bead is placed on the top of the oscillating block just as it reaches its lowest point (see diagram). Assume that the bead has no effect on the frequency or amplitude of the oscillation. (i) At what displacement from the block's equilibrium position does the bead first lose contact with the block? (ii) What is the speed of the bead at the instant when it loses contact with the block?

Explanation / Answer

(i)
The forces on the bead are its weight mg downward and the upward normal force exerted by the block. The magnitude of this normal force changes as the acceleration changes.

Therefore when the acceleration in downward direction is larger than g , the bead will lose contact with the block.

a = - w^2*y
g = - (2*pi*f)^2 * y
y = - 9.8/(2*3.14*4)
y = - 0.0155 m
y = - 1.55 cm
Displacement, y = -1.55 cm

(ii)
Vmax = omega*A
Vmax = 2*pi*4 * 7.0 * 10^-2
Vmax = 1.76 m/s

k = w^2 * m

1/2 *mv^2 + 1/2 kx^2 = 1/2 * m*vmax^2
1/2 *mv^2 + 1/2 w^2 * m *x^2 = 1/2 * m*vmax^2
v^2 + w^2*x^2 = vmax^2
v = sqrt(1.76^2 - (2*pi*4)^2 * 0.0155^2)
v = 1.716 m/s
Speed at which bead leaves the block, v = 1.716 m/s

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