Maximum amplitude of oscillation of a block

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SUMMARY

The maximum amplitude of oscillation for a system consisting of a 26 kg block executing horizontal simple harmonic motion with a frequency of 1.9 Hz and a 9 kg block resting on it is determined by the static friction coefficient of 0.459. The force required to overcome static friction is calculated as 40.52 Newtons, leading to a maximum acceleration of 4.5 m/s² for the smaller block. To ensure the smaller block does not slip, the amplitude must be calculated based on this acceleration limit.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Knowledge of static friction and its coefficient
  • Familiarity with Newton's second law (F=ma)
  • Ability to perform calculations involving mass, force, and acceleration
NEXT STEPS
  • Calculate the maximum amplitude of oscillation using the formula for acceleration in simple harmonic motion
  • Explore the relationship between frequency and amplitude in oscillatory systems
  • Investigate the effects of varying the mass of the blocks on the system's behavior
  • Learn about energy conservation in simple harmonic motion
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Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators looking for practical examples of static friction and harmonic motion applications.

Mac13
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I've been staring at this problem for a loonnng time and i just can't figure it out:(.. i would appreciate it if someone could help me...thx! :)

A large block with mass 26 kg executes
horizontal simple harmonic motion as it slides
across a frictionless surface with a frequency
1:9 Hz : Block smaller block with mass 9 kg
rests on it, and the
coefcient of static friction between the two is
u = 0.459.
The acceleration of gravity is 9:8.
What maximum amplitude of oscillation
can the system have if the block is not to slip?
Answer in units of cm.
 
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The weight of the smaller block is 9(9.81) = 88.29 Newtons. The force necessary to overcome the static friction, then is (88.29)(0.459)= 40.52 Newtons. Using "F= ma", that would correspond to an acceleration (of the smaller) block of 40.52= 9a or
a= 4.50 m/s2. What amplitude would give a maximum acceleration no greater than 4.5 m/s2?
 
thanks...its a little too late to turn it in for my homework because it was due early this morning but at least i understand it now:)
 

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