A sample SHM problem
A block of mass M sits at equilibrium between two springs of spring
constant k_{1} and k_{2}. A bullet of mass m and speed v is
fired horizontally at the block and becomes embedded in it. Thereq is
no friction between the block and the surface.
(a) What is the amplitude of the resulting oscillation?
(b) What is the angular frequency?
How should we connect the motion of the bullet before the
collision to the motion of the block + bullet after the collision?
 Conservation of linear momentum
 Conservation of mechanical energy
 Either one works
Use momentum conservation, because mechanical energy is lost in the
inelastic collision.
Momentum before the collision = momentum after the collision gives:
mv + 0 = (M + m) v_{f} → v_{f} 
= 
m
 
M + m


v 
The block plus bullet has velocity v_{f} at the equilibrium
position, where the potential energy is zero. We now use energy conservation
to find the maximum amplitude: the kinetic energy at the equilibrium position
equals the potential energy stored in the two springs at the maximum
amplitude point.
K_{i}= U_{f} →
½ (m+M)v_{f}^{2} = ½ k_{1}A^{2}
+ ½ k_{2}A^{2}
Cancelling factors of ½ and substituting v_{f} from above
gives:
m^{2} v^{2}
 
M + m


= 
(k_{1} + k_{2}) A^{2} 
This gives an amplitude of 
A 
= 
m v
 
[(M + m)(k_{1} + k_{2})]^{½ }


(b) What is the angular frequency?
There are two ways to solve for ω. One is to remember that for
simple harmonic motion v_{max} = Aω.
Since v_{max} = v_{f}, the speed at the equilibrium position, we have:
ω 
= 
v_{f}
 
A


= 
m v
 
M + m


* 
[(M+m)(k_{1} + k_{2})]^{½}
 
m v


ω 
= 
( 
k_{1} + k_{2}
 
M + m


) 
^{½}

The other solution method is to analyze the forces due to each spring.
Applying Newton's Second Law gives:
ΣF = ma →
k_{1}x  k_{2}x = (M + m) a
a 
=  

k_{1} + k_{2}
 
M + m


x 
For and equation in this form, the angular frequency is the square root of
the factor multiplying  x, so:
ω 
= 
( 
k_{1} + k_{2}
 
M + m


) 
^{½}
