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A 25.0 kg mass is traveling to the right with a speed of 2.80 m/s on a smooth horizontal surface when it collides with and sticks to a second 25.0 kg mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant 170.0 N/m. The other end of the spring is fixed to a wall to the right of the second mass.
1) Find the frequency of the subsequent oscillations.2) Find the amplitude of the subsequent oscillations.3) Find the period of the subsequent oscillations.4) How long does it take the system to return the first time to the position it had immediately after the collision?

Respuesta :

Answer:

Explanation:

1 ) angular frequency ω = √ ( k / m )

=√ ( 170 / 50 )

= 1.844 rad /s

2πn = 1.844 where n is frequency of oscillation

n = 1.844 / (2 x 3.14 )

= .294 per sec

= .294 x 60 = 18 approx. per minute .

Velocity just after collision of composite mass ( using law of conservation of momentum )

= 25 x 2.8 / 50

v = 1.4 m/s

If new amplitude be A

1/2 k A² = 1/2 m v²

m = 25 + 25 = 50 kg

170 x A² = 50 x 1.4²

A = 0.76 m

3 ) period of oscillation = 1 /n

= 1 / .294

= 3.4 s

4 ) It will take complete one period of oscillation ie 3.4 s to come to its original position.

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