What Is the Frequency of Vibration When a Cube Is Released?

[dekoi]

1.) A cube 1.50 cm on edge mounted on the end of strip that lies in vertical plane. Mass of strip is neglible, but the length of the strip is much larger than cube. The other end of strip is clamped on to a stationary frame. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from equilibrium. If the cube is released, what is the frequency of vibration?

Absolutely any hints would be appreciated.


2.) A 2kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 20N is required to hold the object at rest when it is pulled 0.2 m from equil. Object is released from rest with initial position of .2 m . Find:
Force Konstant
Frequency
Max speed.
Max acceleration.
Total Energy
Velocity when position is equal to one third of maximum value is 1.33 m/s.
Acceleration of object when its position is equal to one third of the maximum value.

The questions also had other parts. For these, i found that:
k= 100 N/m
f = 10 Hz
v = 1.4142 m/s
a = 9.99 m/s^2
The velocity when position is equal to one third of maximum value is 1.33 m/s.

However, how do i find the acceleration of object when its position is equal to one third of the maximum value?

Also, it wouldbe nice for someone to check my answers.



Thank you very much.

 

[HallsofIvy]

1.) A cube 1.50 cm on edge mounted on the end of strip that lies in vertical plane. Mass of strip is neglible, but the length of the strip is much larger than cube. The other end of strip is clamped on to a stationary frame. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from equilibrium. If the cube is released, what is the frequency of vibration?

Absolutely any hints would be appreciated.

I don't see any way to give useful hints if you don't tell us what you do understand about this problem or what you have tried. Clearly the "strip" is being considered like a spring. Do you know how to find the "spring constant"? If this were part of a differential equations course, you might then be expected to set up and solve the differential equation of motion. If not then you may have been given a formula for period based on the spring constant.

2.) A 2kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 20N is required to hold the object at rest when it is pulled 0.2 m from equil. Object is released from rest with initial position of .2 m . Find:
Force Konstant
Frequency
Max speed.
Max acceleration.
Total Energy
Velocity when position is equal to one third of maximum value is 1.33 m/s.
Acceleration of object when its position is equal to one third of the maximum value.

The questions also had other parts. For these, i found that:
k= 100 N/m
f = 10 Hz
v = 1.4142 m/s
a = 9.99 m/s^2
The velocity when position is equal to one third of maximum value is 1.33 m/s.

However, how do i find the acceleration of object when its position is equal to one third of the maximum value?

Also, it wouldbe nice for someone to check my answers.

Thank you very much.

It would have been simpler if you had told us what "k", "f", etc. mean!
I guess that "k" is what you called the "frequncy Konstant" and I would call the "spring constant".
Yes, 20 N/.2 m= 100 N/m.
No, I do NOT get 10Hz for the frequency but I do get "1.4142" (√(2)) for the maximum speed. And I get 10 m/s², not "0.99" for the maximum acceleration.

Since you found the velocity when the position is 1/3 of its maximum, I presum that you found the time, t, when that happens (and I would argue that the question is ambiguous- there are two different answers to that question). You should be able to just plug that into the equation for acceleration.

If you could do all that for problem 2, why aren't you able to do problem 1? It's essentially the same.

 

[quicknote]

I would like to provide the following response to the content about oscillatory motion:

1.) The frequency of vibration can be calculated using the formula f = 1/2π * √(k/m), where k is the force constant and m is the mass of the cube. In this case, the force constant can be found by dividing the force applied (1.43 N) by the displacement from equilibrium (2.75 cm or 0.0275 m). This gives a force constant of 52 N/m. The mass of the cube is not given in the question, so we cannot calculate the frequency without that information.

2.) Based on the given information, we can calculate the force constant using the formula k = F/x, where F is the applied force (20 N) and x is the displacement from equilibrium (0.2 m). This gives a force constant of 100 N/m. Using the formula f = 1/2π * √(k/m), we can calculate the frequency of vibration. However, the mass of the object is not given in the question, so we cannot calculate the frequency without that information.

To find the maximum speed, we can use the formula v = Aω, where A is the amplitude (maximum displacement) and ω is the angular frequency, which can be calculated using the formula ω = 2πf. In this case, the amplitude is given as 0.2 m and the frequency can be calculated using the formula f = 1/2π * √(k/m). The maximum speed is then equal to Aω = 0.2 * 2π * f.

To find the maximum acceleration, we can use the formula a = Aω^2, where ω is the angular frequency calculated earlier. The maximum acceleration is then equal to Aω^2 = 0.2 * (2πf)^2.

To find the total energy, we can use the formula E = 1/2kA^2, where k is the force constant and A is the amplitude. In this case, the total energy is equal to 1/2 * 100 * 0.2^2 = 2 J.

To find the acceleration of the object when its position is equal to one third of the maximum value, we can use the formula a = -ω^2x, where ω is the angular frequency