Mass on a Spring Homework: Max Speed, Energy, Acceleration

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Homework Statement


A mass of .4 kg is connected to a spring with a spring constant of 100 N/m, slides on a frictionless horizontal surface in simple harmonic motion, Its maximum displacement is .3m?

What is the maximum speed of the mass?

What is the energy stored in gthe mass/spring system when its speed is half the maximum value?

What is acceleration of t he mass when its speed is half the maximum value?


Homework Equations



m=.4kg
k=100N/m
A(max displacement)=.3m



The Attempt at a Solution



What is the maximum speed of the mass?

Vmax=ωA=√k/m(A)=√100/.4 (.3)=4.74m/s CORRECT?


What is the energy stored in gthe mass/spring system when its speed is half the maximum value?

PE=KE
1/2k(A)2 = 1/2m(v)2 so...
1/2(100)(.3)2 = 1/2(.4)(4.74)
4.50 = 4.50

so...1/2(.4)(4.74/2)2 = 1.12J IS THIS CORRECT?


What is acceleration of t he mass when its speed is half the maximum value?

amax=A(ω)2, but where would I use the speed half max value?

I would really like to learn this. It's been 20+ years since I have taken a Physics course.
Thank you!
 
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ω = 15.8 rad/s , so ωA = 4.745 m/s ; correct.

the PE = KE statement is true for PE average = KE average
(and for PE max = KE max , which formulas you used)
But at any instant (read "stored WHEN its speed ..."), PE + KE = E total.
... when v = ½ v_max , KE = ¼ KE_max . so how much PE is there?
 
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Hi Mom, and welcome to PF. Creative use of the provided template!
(Values for m, k and a are more like "given/known data" than equations)
Just so we're talking about the same problem: The other end of the spring is a fixed point.

"Vmax=ωA=√k/m(A)=√100/.4 (.3)=4.74m/s CORRECT?"
Don't think so. But with a few brackets in the right place and in lower case (all caps is considered shouting in PF and they frown on that...): yes.

We go on with energy stored: no friction means no energy loss. So energy constant. In my humble perception, "energy stored in the mass/spring system" is potential energy from the spring plus kinetic energy from the mass. A constant, also when v=vmax/2.
Value follows from some relevant equation (to be listed under 2...) involving k and xmax.

Story would be different if we were looking at PE only at the point where v=vmax/2. That is not the point where PE=KE for the simple reason that (vmax/2)2= (vmax)2/4 (and not /2).

amax=A(ω)2, but where would I use the speed half max value?
equation is correct, but this time they don't wan amax but the magnitude of a at some specific point in the cycle. List an expression for displacement as a function of Amax, ω and time (under 2. of course...), differentiate once to get the speed and once again to get the acceleration. Find out where in the cycle you are (follows from one of the givens) and substitution gives you a.



Bedtime for me, but someone else will help you further along.
By the way: 20 years isn't that much (I can tell) and from what I see I don't think you'll take long to catch up. Brava! and good luck!
 
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So then would I find E Total first

= 1/2k(A)2 + 1/2 m(v)2
=1/2(100)(.3)2 + 1/2(.4)(4.74)2
E total = 8.99

then subtract value of 1/4m(v)2?
 
Total energy E is constant given by (1/2)kA2 ,where A is the amplitude.