Simple Harmonic Motion and maximum velocity

In summary, the conversation discusses two problems related to simple harmonic motion. The first problem involves finding the spring constant of spring 2 given the spring constant and amplitude of spring 1. The second problem involves determining the amount of spring stretch when a ball is attached to one end of a spring and whirled around in a horizontal circle. The conversation also includes equations and attempts at solving the problems, with the final solution being 696N/m for the first problem and 2.29 x 10^-3 for the second problem.
  • #1
kiro626
4
0
[SOLVED] Simple Harmonic Motion

I'm really having a hard time solving these two problems.

1. Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is 174 N/m. The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring 2. The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2.

Homework Equations


the relevant equations are those for SHM. F=kx and f=-kx.

The Attempt at a Solution


I tried the formulas but I still can't find the answer. The answer is supposed to be 696N/m
2nd problem:

A small ball is attached to one end of a spring that has an unstrained length of 0.200 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.00 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.010 m. By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Homework Equations


same as the one above

The Attempt at a Solution


Again with this one, I tried solving it using the same formula but I can't figure how to use the velocity since its in m/s rather than rad/s for uniform circular motion. The answer is supposed to be 2.29 x 10^-3

Any help would be gladly appreciated
 
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  • #2
This is quite a neat problem. The position of the spring is given by [itex] y = A sin(\omega t) [/itex] where [itex] A [/itex] is the amplitude and [itex] \omega = \sqrt{\frac{k}{m}} [/itex]. You need to work out what the velocity will be and then set them equal. REmember the max velocity will be at the equilibrium point.
 
  • #3
I already solved the first problem but I still can't figure the second one out. I tried equating the constant of spring since they would be equal but it give me 2.18x10^-3 so its a little off
 
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  • #4
WEll if you show your full work it would help me. I suspect you're using the wrong value for r.
 
  • #5
uhmm... for the second problem i used the equation:

ma = kx

wherein a is centripetal acceleration
i used 0.2 as the r and then 0.01 as the x to get the k then i equated the k from the second equation that i made which is:

mg = kx(sub2)

I'm not really sure about it
 
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  • #6
it shall do a lot of good if you tell us why are you not sure?
 
  • #7
kiro626 said:
uhmm... for the second problem i used the equation:

ma = kx

wherein a is centripetal acceleration
i used 0.2 as the r and then 0.01 as the x to get the k then i equated the k from the second equation that i made which is:

mg = kx(sub2)

I'm not really sure about it

I'd have used 0.21 as r.
 
  • #8
Kurdt said:
I'd have used 0.21 as r.

thanks I finally got the answer. I finally understood it ^^ thanks very much.
 

Related to Simple Harmonic Motion and maximum velocity

1. What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion where an object oscillates back and forth around a central equilibrium point due to a restoring force that is directly proportional to the displacement from the equilibrium point.

2. What is maximum velocity in SHM?

Maximum velocity in SHM refers to the highest speed that an object reaches during its oscillations. This occurs at the equilibrium point where the restoring force is zero, and the object has the most kinetic energy.

3. How is maximum velocity related to amplitude in SHM?

The maximum velocity in SHM is directly proportional to the amplitude of the oscillations. This means that as the amplitude increases, the maximum velocity also increases, and vice versa.

4. Can maximum velocity be negative in SHM?

Yes, maximum velocity can be negative in SHM. This occurs when the object is moving away from the equilibrium point and its velocity is in the opposite direction of the restoring force. However, the magnitude of the velocity will still be the same as the maximum velocity at the equilibrium point.

5. How can the maximum velocity of an object in SHM be calculated?

The maximum velocity of an object in SHM can be calculated using the equation v = ωA, where v is the maximum velocity, ω is the angular frequency, and A is the amplitude of the oscillations. It can also be calculated using the equation v = √(k/m)A, where k is the spring constant and m is the mass of the object.

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