Question 136: Oscillations

Question

An oscillator of mass M is at rest in its equilibrium position in a potential 
V = $${1 \over 2}$$ k(x $$-$$ X)2.  A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it.  This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is : (M = 10, m = 5,  u = 1,  k = 1)

Accepted Answer

Potential of the given oscillator is
$$V = {1 \over 2}k{(x - k)^2}$$
Given: M = 10; m = 5, u = 1; k = 1
Initial momentum of the particle of mass m
= mu = m $$	imes$$ 5 = 5m
Momentum of (oscillator + particle) after collision = (M + m)
Velocity of oscillator after collision = v
So, momentum of system = (M + m)v
From conservation of linear momentum, we have
(M + m) = mu = 5 $$	imes$$ 1 = 5
For second collision, oscillator and particle have momentum in opposite direction.
Net or total momentum is zero.
Likewise after 4
th
, 6
th
, 8
th
, 10
th
, 12
th
collision the momentum is zero. After 12
th
collision, Mass of oscillator and 12 particles will be (10 + 12 $$	imes$$ 5) = 70
Now, from conservation of linear momentum, for 13
th
collision, we have
$$70 	imes 0 + 5 	imes 1 = (70 + 5)v' \Rightarrow v' = {5 \over {75}} \Rightarrow {1 \over {15}}$$
Total mass after 13
th
collision = (10 + 13 $$	imes$$ 5) = 75
Kinetic energy of system $$ = {1 \over 2}mv{'^2}$$
$$ \Rightarrow KE = {1 \over 2} 	imes 75 	imes {1 \over {15}} 	imes {1 \over {15}}$$
$$ \Rightarrow {1 \over 2}k{A^2} = {1 \over 2} 	imes {{75} \over {225}} = {1 \over 6}$$
$$ \Rightarrow {1 \over 2} 	imes 1 	imes {A^2} = {1 \over 6} \Rightarrow {A^2} = {1 \over 3}$$
$$ \Rightarrow A = {1 \over {\sqrt 3 }}$$

Answer

$${1 \over {\sqrt 3 }}$$

Answer

$${1 \over 2}$$

Answer

$${2 \over 3}$$

Answer

$${3 \over {\sqrt 5 }}$$

Oscillations

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