Just need Simple Harmonic Motion problem checked

In summary, when the block collided with the spring, it compressed the spring 4m from its rest position in 3.4N of force.
  • #1
phys1618
106
0

Homework Statement


a 1kg block sliding a frozen surfaces collideds w/a horizontal spring w/a force constant of 2N/m. The block compresses the spring 4m fr. its rest position.
a.how long does it take to fully colmpress the spring?b. what wa sthe speed of the block at the instant of the collision?c. when the springs is compressed 1.7m, how much force is the block on it?


Homework Equations


k=force constant A=4m x=1.7m
T=2pi(sqroot (m/k)
vmax=(2piA)/(T)
F/x=k


The Attempt at a Solution


I used the T eq. to find the answr for a. which is 4.44s
Then I took T and plugged in for vmax eq. and have the speed as 5.66m/s
for c. I used the f/x=k equation and came up with F=3.4N

is this the way I'm suppose to do it, it somehow seems too easy to be true? =x
 
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  • #2
Your answer for a is wrong. And since you used your answer for a to answer b, b is wrong too. You used the equation of the period to find the time to fully compress the spring. But the period is the time required to execute a full cycle, not compress the spring once. You got c right.
 
  • #3
can you hint me because i can't find an another equation with T in it. =x sorry...
 
  • #4
He doesn't want you do use another equation, I reckon he wants you to understand something conceptually. T=period, and if you have an oscillating system and start at x=A, then the time it takes to go from there and back to the exact same position x=A is the period, T. So, a full cycle would be a full oscillation. If you go from x=A to x= -A, it would be ½T, meaning half an oscillation. The time you want to find is neither--your compression isn't a full cycle (which is what you've calculated) or ½ a cycle. You go from x=0 to x=A, which is how much of a full oscillation?
 
  • #5
1/T? sorry, but I am a lil confused because a full oscillation is when the block hitted the spring and goes back to its original state right? so when it was compressed, it would be half of the compressed so its 1/4 of the T equation?
 
  • #6
a full oscillation is when the block hitted the spring and goes back to its original state right? so when it was compressed, it would be half of the compressed so its 1/4 of the T equation?

Actually, as you go from rest to amplitude and back to rest again you only do ½ an oscillation--a full one is when you're at the same position with the velocity vector pointing the same way. When you go back to the rest position, the velocity is backwards (so to speak). So, the block would have to reach -A (meaning the full extension) and back again to the rest position to complete one oscillation. So, you have only a quarter of an oscillation. (Am I even making sense? it's late where I live and I'm tired.)

But on the other hand, you are correct when you say it's 1/4 T.
 
  • #7
Yes, compressing the spring from its equilibrium position takes 1/4 of a period.
 
  • #8
thank you for taking your time to help me. I am sorry for keeping you up.

Imma try the problem again. and hopefully i'll get it right this time thank you.
 
  • #9
so i used (1/4)T=2pi(sqroot(m/k) and get 1.11s? then use this inthe velocity equation and get v= 22.64m/s?
thanks for the help!
 
  • #10
is this correct?
 
  • #11
Yeah, I'd say that's correct.
 
  • #12
thank you so much for the help..i love this site... you all are so helpful and helps me understand the proble,m more...thank you all for the help...
 

Related to Just need Simple Harmonic Motion problem checked

1. What is Simple Harmonic Motion?

Simple Harmonic Motion is a type of periodic motion where an object moves back and forth in a straight line with a constant frequency and amplitude. This type of motion is commonly found in systems with restoring forces, such as springs and pendulums.

2. How do I solve a Simple Harmonic Motion problem?

To solve a Simple Harmonic Motion problem, you need to identify the variables given, such as amplitude, frequency, and mass, and use the equations of motion to calculate the unknown variables. These equations include the period, frequency, and amplitude formulas.

3. What is the difference between Simple Harmonic Motion and Uniform Circular Motion?

The main difference between Simple Harmonic Motion and Uniform Circular Motion is the type of path the object follows. In Simple Harmonic Motion, the object moves back and forth along a straight line, while in Uniform Circular Motion, the object moves along a circular path.

4. What is the relationship between Simple Harmonic Motion and Hooke's Law?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This relationship is also observed in Simple Harmonic Motion, where the restoring force is proportional to the displacement from the equilibrium position.

5. How is Simple Harmonic Motion used in real-life applications?

Simple Harmonic Motion is used in many real-life applications, such as springs in car suspensions, the swinging motion of a pendulum in a clock, and the vibrations of guitar strings. It is also used in the study of earthquakes and other natural phenomena.

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