Maximum acceleration, frequency and mechanical energy of a spring

In summary, when a 0.20-kg block is suspended from a vertically hanging spring and then attached to the same spring on a horizontal surface, it undergoes simple harmonic motion with a maximum acceleration of 98m/s, a frequency of √(k/m)/2∏, and a total mechanical energy that is constant and equal to the sum of potential and kinetic energy. The amplitude of motion is 0.05m.
  • #1
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1. When a 0.20-kg block is suspended from a vertically hanging spring, it stretches the spring from its original length of 0.050m to 0.060m. The same block is attached to the same spring and placed on a horizontal, friction-less surface. The block is then pulled so that the spring stretches to a total length of 0.10m. The block is released at time t=0 s and undergoes simple harmonic motion.

0.20 kg
[_]-/\/\/\/\|
________spring

1. What is the maximum acceleration of the block?
2. What is the frequency of the motion?
3.What is the total mechanical energy of the system at any instant?F=mg=-k/x
a=(-k/m)x
ω=2∏/T
Emechanical energy=U+k
For acceleration it should be 98m/s because when the mass was hanging down the spring x=.010m and F=1.96 that would mean -k=196 then plugged into a=-k/m*x with .1 as x the answer should be 98m/s. I'm sure it's right, but let me know if I'm wrong. I'm having trouble though finding the other two answers, for frequency of motion I can't find time and for mechanical energy I can't find the equation for potential energy in a spring.
 
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  • #2
Mechanical energy (PE+KE) is constant in this system. That being the case, the derivative of it with respect to time is zero. Assume the displacement to be simple harmonic motion such as

x = A*sin(omega*t)

Knowing that PE=(kx^2)/2 and KE=(mv^2)/2, you can proceed from there to get the natural frequency.
 
  • #3
But how would you get velocity?
 
  • #4
The velocity at any time is simply the time derivative of the position.
 
  • #5
I found v=+ or -√(k/m)√A^2 -x^2 with A equals the amplitude of motion but what's the amplitude of motion?
 
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  • #6
"v=+ or -√(k/m)√A^2 -x^2"

Where did you get the above expression? The -x^2?

The amplitude of the motion is given in the problem statement.
 
  • #7
It was in my physics book, the square root of k over m times the square root of A squared minus x squared. Is the amplitude of motion the same as x, .1 or is it .01? I just forgot to put it into parentheses.
 
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  • #8
"its original length of 0.050m ...The block is then pulled so that the spring stretches to a total length of 0.10m."

Problem states that the original length is 0.05m. So if it is initially streached to 0.10m, the amplitude is 0.05m.
 

1. What is maximum acceleration of a spring?

The maximum acceleration of a spring refers to the highest acceleration value that a spring can reach when it is fully compressed or extended. This value is dependent on the spring's stiffness, mass, and the force applied to it.

2. How is frequency related to a spring's mechanical energy?

The frequency of a spring refers to the number of oscillations it can complete in one second. It is directly proportional to the spring's mechanical energy, meaning that as the frequency increases, the mechanical energy also increases. This is because a higher frequency requires the spring to oscillate faster, resulting in more mechanical energy being stored and released.

3. How can the mechanical energy of a spring be calculated?

The mechanical energy of a spring can be calculated using the equation E = 1/2kx^2, where E is the mechanical energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This equation takes into account the potential energy stored in the spring when it is compressed or extended.

4. What factors affect the maximum acceleration of a spring?

The maximum acceleration of a spring is affected by its stiffness, mass, and the force applied to it. A stiffer spring will have a higher maximum acceleration, while a heavier mass and a greater force will also result in a higher maximum acceleration.

5. Can the frequency of a spring be changed?

Yes, the frequency of a spring can be changed by altering its stiffness or mass. A stiffer spring will have a higher frequency, while a heavier mass will have a lower frequency. Additionally, the frequency can also be changed by altering the force applied to the spring, as a greater force will result in a higher frequency.

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