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

(1)

The angular velocity is,

=(m/k )=(g/A)

=(9.8/0.098)

= 10

The time period is,

T = 2 / 10

= / 5

= 0.628 s

(b)

The acceleration is,

a = (2f)2A

=(2(5/))2(0.098)

= 9.8 m/s2
since its going down, the a = -9.8 m/s2

(c) the force is

F = ma

But at equilibrium position, the acceleration is zero. So, the net force is zero.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.