183_notes:examples:energy_in_a_spring-mass_system

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A mass of 0.2 kg is attached to a horizontal spring whose stiffness is 12 N/m. Friction is negligible. At t = 0 the spring has a stretch of 3 cm and the mass has a speed of 0.5 m/s.

a: What is the amplitude (maximum stretch) of the oscillation?

b: What is the maximum speed of the block?

Facts

a:

Initial State: s_i = 3 cm, v_i = 0.5 m/s

Final State: v_f = 0 (maximum stretch)

b:

Initial State: s_i = 3 cm, v_i = 0.5 m/s

Final State: Maximum speed (s_f = 0)

Lacking

Approximations & Assumptions

Representations

System: Mass and spring

Surroundings: Earth, table, wall (neglect friction, air)

Solution

a:

From the Energy Principle:

Ef=Ei+W

Kf+Uf=Ki+Ui+W

Kf and W cancel out.

0+12kss2f=12mv2i+12kss2i

12kss2f=12(0.2kg)(0.5m/s)2+12(12N/m)(0.03m)2

12kss2f=0.03J

sf=2(0.03J)/(12N/m)=0.07m

sf=0.07m

b:

Kf+0=0.03J

12mv2f=0.03J

vmax=2(0.03J)/(0.2kg)=0.55m/s

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