Simple Harmonic Motion (Bullet fired into a block attached to spring)

In summary: So, you want to use the maximum compression, A=.15m and the mass of the block+bullet, .505kg. The speed will be the same as the velocity when the block and bullet are moving together.In summary, the conversation discusses a problem involving a bullet being fired horizontally into a block of wood attached to a spring. The system goes into simple harmonic motion with a frequency of 9Hz and amplitude of 15cm. The speed of the bullet can be determined by using the equation for conservation of energy, which requires finding the spring constant and using the maximum compression and total mass of the system. The resulting speed is 0.1047m/s.
  • #1
fluffymastr
1
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Homework Statement



A 5g bullet is fired horizontally into a 0.50kg block of wood resting on a frictionless table. The block, which is attached to a horizontal spring, retains the bullet and moves forward, compressing the spring. The block-bullet system goes into SHM with a frequency of 9Hz and amplitude of 15cm.

A) Determine the speed of the bullet


Homework Equations



Fs=-kx
a=-(k/m)x
PEs= .5kx2
v=±[(k/m)(A2-x2)].5
T=2π(m/k).5

The Attempt at a Solution



I wasn't sure how to even approach this but I think I need to determine the spring constant k using a form of T=2π(m/k).5.

So:
T=2π(m/k).5
T2=4π2(m/k)
T2=4π2(m/k)
k=(4π2m)/(T2)
k=(4π2(.5kg+.005kg))/(92)
k=.24613kg/s2

Then plug that into:
v=±[(k/m)(A2-x2)].5
v=±[(.24613/.505)(.152-02)].5
v=.1047m/s

I have NO IDEA if I am right or not. Anyone want to confirm or deny?
 
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  • #2
T=2π(m/k).5
T2=4π2(m/k)
T2=4π2(m/k)
k=(4π2m)/(T2)
k=(4π2(.5kg+.005kg))/(92)
k=.24613kg/s2

The frequency is 9 Hz, not the period.
Then plug that into:
v=±[(k/m)(A2-x2)].5
v=±[(.24613/.505)(.152-02)].5
v=.1047m/s

I have NO IDEA if I am right or not. Anyone want to confirm or deny?

That would be right if you used the right value of k. The equation comes from the conservation of energy, which says that 1/2*k(A^2-x^2)=1/2mv^2.
 

Related to Simple Harmonic Motion (Bullet fired into a block attached to spring)

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium point in a straight line. The motion is characterized by a constant amplitude and a sinusoidal velocity or acceleration.

2. How is a bullet fired into a block attached to a spring an example of simple harmonic motion?

When a bullet is fired into a block attached to a spring, the block will begin to oscillate back and forth due to the energy transferred from the bullet's impact. The block's motion will follow a sinusoidal pattern, making it an example of simple harmonic motion.

3. What factors affect the frequency of simple harmonic motion?

The frequency of simple harmonic motion is affected by the mass of the object, the spring constant of the spring, and the amplitude of the oscillation. Increasing the mass or decreasing the spring constant will decrease the frequency, while increasing the amplitude will increase the frequency.

4. How does the amplitude of the oscillation affect the energy of simple harmonic motion?

The amplitude of the oscillation directly affects the maximum potential and kinetic energy of the system. As the amplitude increases, so does the maximum energy. This is because the potential energy stored in the spring and the kinetic energy of the object increase as the amplitude increases.

5. Is simple harmonic motion a real-life phenomenon?

Yes, simple harmonic motion can be observed in many real-life situations, such as the swinging of a pendulum, the vibrations of a guitar string, or the motion of a mass attached to a spring. It is also used in various technologies, such as clocks and shock absorbers.

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