Maximum acceleration, frequency and mechanical energy of a spring

by jmgXD6
Tags: frequency of motion, maximum acceleration, mechanical energy, spring
jmgXD6 is offline
May8-12, 01:44 PM
P: 17
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

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?

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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LawrenceC is offline
May9-12, 01:59 PM
P: 1,195
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.
jmgXD6 is offline
May9-12, 07:47 PM
P: 17
But how would you get velocity?

Steely Dan
Steely Dan is offline
May9-12, 09:56 PM
P: 317

Maximum acceleration, frequency and mechanical energy of a spring

The velocity at any time is simply the time derivative of the position.
jmgXD6 is offline
May10-12, 08:14 AM
P: 17
I found v=+ or -√(k/m)√A^2 -x^2 with A equals the amplitude of motion but what's the amplitude of motion?
LawrenceC is offline
May10-12, 09:18 AM
P: 1,195
"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.
jmgXD6 is offline
May10-12, 09:32 AM
P: 17
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.
LawrenceC is offline
May10-12, 10:40 AM
P: 1,195
"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.

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