Question

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass 1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional). Suddenly, Slimer* holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.48 kg.
Required:
Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.

Answer 1

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m