Simple harmonic motion - answer slightly off

In summary, a 4.00kg block suspended from a spring with k = 500N/m has a 50.0g bullet fired into it from directly below with a speed of 125m/s, becoming embedded in the block. Using the conservation of momentum and the equation for kinetic energy, the percentage of the initial kinetic energy of the bullet transferred to mechanical energy of the oscillator is found to be 1.17%. This differs from the book's answer of 1.23% due to early rounding.
  • #1
Lord Anoobis
131
22

Homework Statement


A 4.00kg block is suspended from a spring with k = 500N/m. A 50.0g bullet is fired into the block from directly below with a speed of 125m/s and becomes embedded in the block.

a) Find the amplitude of the resulting SHM.
b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator?

Homework Equations


Conservation of momentum, ##p_f = p_i##
##K = \frac{1}{2}mv^2##

The Attempt at a Solution


The problem is solved but there is something about the given answer or part (b) that is bugging me, so let's ignore (a). For (b) I arrived at

##percentage = 100(\frac{K_m}{K_b})##

##percentage = 100(\frac{m(v_b)^2}{(m + M)(v_m)^2})##

where ##K_b## is the initial kinetic energy of the bullet and ##K_m## is the kinetic energy of system after the bullet becomes embedded in the block. The velocity of the block in this state is given by

##v_m = \frac{m v_b}{m + M}##

The trouble is that the preferred method of working with symbols until the very end gives an answer of 1.23% whereas the book answer is 1.17%. However, I noticed that if you use ##v_m = 1.5m/s## the correct answer pops up.

I've noticed this sort of thing on a number of occasions where using a rounded figure in the final answer turns out to be more correct, so to speak, which goes against the usual method. Did I blow it somewhere or is it really a case of early rounding?
 
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  • #2
Book answer is simply wrong -- due to too early rounding, as you indicate. Well spotted !
 
  • #3
BvU said:
Book answer is simply wrong -- due to too early rounding, as you indicate. Well spotted !
Thanks a lot. Always good to know where the error lies.
 

Related to Simple harmonic motion - answer slightly off

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 with a constant amplitude. This type of motion is characterized by a restoring force that is directly proportional to the displacement of the object from the equilibrium point.

2. What are some examples of simple harmonic motion?

Some examples of simple harmonic motion include a pendulum swinging back and forth, a mass-spring system vibrating up and down, and the motion of a mass attached to a rotating wheel.

3. How is simple harmonic motion different from other types of motion?

Simple harmonic motion is different from other types of motion because it is a periodic motion with a constant amplitude and a restoring force that is proportional to the displacement. Other types of motion may have varying amplitudes or non-proportional restoring forces.

4. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A*cos(ωt + φ), where x is the displacement of the object, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

5. What factors affect the period of simple harmonic motion?

The period of simple harmonic motion is affected by the mass of the object, the stiffness of the restoring force, and the amplitude of the motion. It is also affected by the gravitational acceleration and the length of the pendulum or spring in the system.

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