A block of mass m rests on a horizontal platform. The platform is driven vertica

Question

A block of mass m rests on a horizontal platform. The platform is driven vertically in a simple harmonic motion with an amplitude of 0.098m. When at the top of its path, the block just leaves the surface of the platform. ( this means that at this point its acceleration is 9.8ms-2 downward).

1) what is the period of the simple harmonic motion?

2) when the block is at the bottom point of its path, what is its acceleration?

3) what is the force exerted by the platform on the block at this bottom point?

i would detailed answers to the questions please

Explanation / Answer

let w is the angular frequecy of motion.

1) a_max = A*w^2

g = A*w^2

==> w = sqrt(g/A)

= sqrt(9.8/0.098)

= 10 rad/s

Time period, T = 2*pi/w

= 2*pi/10

= 0.628 s

2) at lowest point, Fnet = F_spring - m*g

m*a = F_spring - m*g

2*m*g = m*a

a = 2*9.8

= 19.6 m/s^2

3) Let F is the force exerted by the platform on the block.

Fnet = F - m*g

m*a = F - m*g'

F = m*g + m*a

= m*g + m*g

= 2*m*g

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