# Maximum acceleration, frequency and mechanical energy of a spring

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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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.

But how would you get velocity?

The velocity at any time is simply the time derivative of the position.

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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"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.

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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"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.