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

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SUMMARY

The discussion centers on a physics problem involving simple harmonic motion (SHM) where a 5g bullet is fired into a 0.50kg block attached to a spring, resulting in a system with a frequency of 9Hz and an amplitude of 15cm. The user calculated the spring constant (k) to be approximately 0.24613 kg/s² using the formula T=2π(m/k) and subsequently determined the bullet's speed to be approximately 0.1047 m/s. The calculations were confirmed to be correct when applying the conservation of energy principle, equating potential energy and kinetic energy in the system.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with spring constant calculations
  • Knowledge of energy conservation principles in physics
  • Ability to manipulate equations involving mass, frequency, and amplitude
NEXT STEPS
  • Study the derivation of the spring constant (k) in SHM systems
  • Learn about energy conservation in mechanical systems
  • Explore the relationship between frequency and period in oscillatory motion
  • Investigate the effects of mass on the frequency of SHM
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to reinforce concepts of simple harmonic motion and energy conservation.

fluffymastr
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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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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.
 

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