Simple Harmonic motion and collisions

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SUMMARY

The discussion focuses on the dynamics of simple harmonic motion (SHM) and collisions involving two blocks with masses of 2 kg and 4 kg. The 2 kg block oscillates with a period of 20 ms, described by the equation x=(1.0 cm)cos(wt+2π). After a completely inelastic collision at t=5.0 ms, the final velocity of the combined mass is determined to be 4 m/s. To find the amplitude of oscillation post-collision, participants emphasize the conservation of momentum and the relationship between maximum velocity and amplitude in SHM.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Knowledge of momentum conservation principles
  • Familiarity with the equations of motion for oscillating systems
  • Ability to calculate angular frequency (ω) and spring constant (k)
NEXT STEPS
  • Calculate the amplitude of oscillation after the collision using conservation of momentum.
  • Determine the new angular frequency (ω) of the resulting SHM after the collision.
  • Explore the relationship between maximum velocity and amplitude in SHM.
  • Investigate the effects of mass changes on the frequency of oscillation in SHM.
USEFUL FOR

Physics students, mechanical engineers, and anyone studying dynamics and oscillatory motion will benefit from this discussion.

cherylsc
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A block of mass 2kg oscillates on the end of a spring in SHM with a period of 20ms. The position of the block is given by: x=(1.0cm)cos(wt+2pi). Another block of mass 4kg slides toward the oscillating block with velocity of 6m/s. The two blocks undergo a completely inelatic collision at time t=5.0ms.

a)what is the amplitude of oscllaltion after collision?
I know momentum is conserved in the collision and time t=.005 secons, the velocity of the oscillating block is O. So I got V(final)=4m/s.
But I am not sure how to account for this with the spring force to get the final amplitude.

b) What is the frequency of the resulting SHM?
without A, I don't think I can do this part, so I haven't tried yet...
 
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cherylsc said:
A block of mass 2kg oscillates on the end of a spring in SHM with a period of 20ms. The position of the block is given by: x=(1.0cm)cos(wt+2pi).

Are you sure it's cos(wt+2pi)? Might as well make it cos(wt).

Another block of mass 4kg slides toward the oscillating block with velocity of 6m/s. The two blocks undergo a completely inelatic collision at time t=5.0ms.

a)what is the amplitude of oscllaltion after collision?
I know momentum is conserved in the collision and time t=.005 secons, the velocity of the oscillating block is O. So I got V(final)=4m/s.

If the position is given by:
x=(1.0cm)cos(wt+2pi) then, t=5.0ms corresponds to an angle of pi/2. The velocity equation is:
v=-w(1.0cm)sin(wt+2pi). v does not equal 0 at t=5.0ms.

I'd do part b first to get the new w... then use the fact that the magnitude of the maximum velocity = wA. Since you have the maximum velocity (you get it from the conservation of momentum etc...) you can get A (amplitude).

But I am not sure how to account for this with the spring force to get the final amplitude.

b) What is the frequency of the resulting SHM?
without A, I don't think I can do this part, so I haven't tried yet...

w=sqrt(k/m).

You can calculate k, then solve for the new w using a mass of 6kg
 
oops

I meant (1/2)pi

thanks
 

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