Simple Harmonic Motion of a gun

In summary, the conversation discusses the calculation of the speed at which a 0.14 kg ball will leave a toy popgun when compressed with a force of 75 N. The equation F=kx is used to solve for K, which is determined to be 416 N/m. The concept of spring energy and its conversion into kinetic energy is also mentioned, with the formula PE=1/2kx^2 being used to find the potential energy. It is ultimately determined that the velocity of the ball can be calculated using the equation KE=1/2mv^2, with the initial force of 75 N being used to find K. The person asking for help initially makes a mistake but is corrected by the others in the
  • #1
peaches1221
15
0
(1)It takes a force of 75 N to compress the spring of a toy popgun 0.18 m to "load" a 0.14 kg ball. With what speed will the ball leave the gun?

I used the equation F=kx
I solved for K and I got 416 N/m

Then I used the equation omega = square root(K/M)
I solved for omega and i got 54.5.

Is this correct? It seems wrong to me. Please help, any suggestions will be greatly appreciated.
 
Last edited:
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  • #2
Hint: This is not a S.H.M. problem.
f= -k*x
k>> 416 N/m
This is not necessary.
 
  • #3
The spring energy is converted into kinetic energy. Do you know what the spring energy is now that you know K? Can you figure out the velocity with this?
 
  • #4
Thank you both for your help. I'm still not sure if I completely understand this problem

PE= 1/2kx^2
where I know k because it is 416 N/M and X=.18. so I would solve for the PE.

Then you're saying if the spring energy is converted to kinetic energy, then is KE=PE?
thus KE= 1/2mv^2 and then I would have to solve for v?

But I still don't think this is right because where is the 75N coming in?
 
  • #5
You used it to find K, did you not?
 
  • #6
Yeah, I'm an idiot. Thanks for your help.
 

Related to Simple Harmonic Motion of a gun

1. What is simple harmonic motion (SHM) of a gun?

Simple harmonic motion refers to the back-and-forth motion of an object around a central equilibrium point, caused by a restoring force that is directly proportional to the displacement from that point. In the case of a gun, the SHM is the motion of the barrel as it recoils back and forth after firing a bullet.

2. What factors affect the SHM of a gun?

The SHM of a gun can be affected by a number of factors, including the mass of the gun, the force of the bullet being fired, and the stiffness of the gun's recoil spring. The angle at which the gun is held and the surface it is fired on can also impact the SHM.

3. How does the SHM of a gun impact its accuracy?

The SHM of a gun can have a significant impact on its accuracy. If the gun is not held steady and the SHM is not consistent, the bullet's trajectory can be affected, leading to a less accurate shot. This is why it is important for shooters to have a stable grip and stance when firing a gun.

4. Can the SHM of a gun be modified or controlled?

Yes, the SHM of a gun can be modified or controlled through various means. For example, the recoil spring can be adjusted to change the stiffness and therefore the recoil velocity of the gun. Additionally, adding weights to the gun can also affect its SHM.

5. How is SHM used in gun design and engineering?

SHM is an important concept in gun design and engineering. It is used to ensure that the gun's recoil is manageable and that the gun is accurate and reliable. Engineers must carefully consider the SHM of a gun when designing its components, such as the recoil spring and barrel, to ensure that it functions properly and safely.

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