A block of mass M sits at equilibrium between two springs of spring constant k₁ and k₂. A bullet of mass m and speed v is fired horizontally at the block, becoming embedded in it.

(a) What is the amplitude of the resulting oscillation?

(b) What is the angular frequency?

Momentum before the collision = momentum after the collision.

mv + 0 = (M + m) v_f

vf

=

m M + m

v

The block plus bullet has this velocity at the equilibrium position, where the potential energy is zero. We can now use conservation of energy to find the amplitude - the kinetic energy at the equilibrium position equals the potential energy stored in the two springs at the end-point of the motion, where the mass comes instantaneously to rest.

K_i = U_f

½ (m+M)v_f² = ½ k₁A² + ½ k₂A²

Cancelling factors of ½ and substituting our expression for v_f from above gives:

m2 v2 M + m

=

(k1 + k2) A2

This gives an amplitude of

A

=

m v [(M + m)(k1 + k2)]½

(b) What is the angular frequency?

There are two ways to approach this. One is simply to remember that for simple harmonic motion v_max = Aω.

v_max = v_f, the speed at the equilibrium position, so:

ω

=

vf A

=

m v M + m

*

[(M+m)(k1 + k2)]½ m v

ω

=

(

k1 + k2 M + m

)

½

The other way to approach this is to analyze the forces. Each spring applies a restoring force to the system, so applying Newton's Second Law gives:

ΣF = ma

-k₁x - k₂x = (M + m) a

a

= -

k1 + k2 M + m

x

When we get the equation in this form the angular frequency is the square root of whatever is multiplying -x, so:

ω

=

(

k1 + k2 M + m

)

½