Oscillation amplitude of block

In summary, we have a system of two connected blocks on a frictionless surface with a spring attached to a wall. The maximum amplitude of oscillation without the top block slipping on the bottom block can be found by setting the force of the spring equal to the maximum static friction force. Using the equations F=ma, a(t)=-w2Acos(wt+FI), w=sqrt(k/m), and Fs=u*normal force, we can derive an equation to solve for the maximum amplitude A=0.056m. However, the correct answer is 0.1764m, so further assistance may be needed to find the solution.
  • #1
Badmouton
7
0
1.A block of mass M=1 kg rests on a frictionless surface and is connected to a horizontal spring of force constant k=25 N/m. The other end of the spring is attached to a wall. A second block of mass m=500 g, rests on top of the first block. The coefficient of static friction between the blocks is 0.3. Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.



2.These are the equations I used:
F=ma
a(t)=-w2Acos(wt+FI)
w=sqrt(k/m)
Fs=u*normal force




3. Since we don't want the upper block to move, then the force acting on it, namely the spring bobbing under it, must be lower or equal to Fs. This means that the mass*a=Fs. Since I already have the value of the mass (M+m=1kg+0.5g=1.5kg), all I have to find is the acceleration. I calculated the value of w to be 4.08 and the value of Fs to be 1.4. With the acceleration found the following way F=Fs, I derived 1.5kg*a=1.4, thus a=0.93, I can us this equation to isolate A "a(t)=-w2Acos(wt+FI)", since I want the max acceleration, I will assume cos(wt+FI)=1.
Finally this leads me to this equation 0.93=-(4.08)2A, thus A=0.056m... However, the right answer is 0.1764m, can someone help?

 
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  • #2
Fs is proportional to the normal force between the blocks. What is it?

ehild
 
  • #3
It would be the mass of the upper block times gravity---> 9.8N/Kg*0.5kg=4.9N
 
  • #4
The upper block must accelerate with a=Aw2. The force exerted on it is Fs. So Fs=(0.5 kg)*a.

ehild
 
  • #5


I would first commend the individual for their use of equations and calculations in attempting to solve the problem. However, I would also suggest that they double check their calculations and equations to ensure accuracy. It is possible that a mistake was made in the derivation or calculation process, leading to the incorrect answer. Additionally, I would suggest trying different methods or approaches to the problem to see if a different solution is obtained. It is also important to consider any assumptions made and whether they are valid in this scenario. Overall, it is important to carefully review and analyze the problem and its components to reach a correct and accurate solution.
 

FAQ: Oscillation amplitude of block

What is the definition of oscillation amplitude?

The oscillation amplitude of a block is the maximum distance that the block moves away from its equilibrium position during one complete oscillation cycle.

How is oscillation amplitude related to the mass and stiffness of the block?

The oscillation amplitude is directly proportional to the mass of the block and inversely proportional to the stiffness of the block. This means that a heavier block or a less stiff block will have a larger oscillation amplitude.

What factors can affect the oscillation amplitude of a block?

The oscillation amplitude of a block can be affected by the mass, stiffness, and damping of the block, as well as the frequency and amplitude of the driving force.

How does the oscillation amplitude change over time?

The oscillation amplitude of a block will decrease over time due to the effects of damping. This means that the block will eventually come to rest at its equilibrium position.

Can the oscillation amplitude of a block be greater than its initial displacement?

No, the oscillation amplitude of a block cannot be greater than its initial displacement. The initial displacement sets the limit for the maximum oscillation amplitude that the block can achieve.

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