- #1
sm1t
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Hi, Been registered for a while here, but this is my first post, been using the forum as more of a resource before. I am going through some past papers, but I am faltering at 1 question.
A particle moving along the x-axis with velocity v experiences a resistive force –bv, but no spring-like restoring force, where b is the friction coefficient.
(a) Write down the equation of motion. [3]
\ddot{x}= -(b/m)v
(b) Show that the equation of motion is satisfied by
x(t) = C - (v_{0}/\gamma)e^{-\gamma}
where , m is the mass of the particle, and C and \gamma=(b/m) are free parameters. [3]
Not the quickest with latex so the answer was to just differentiate twice, you can then see they are equivalent.
(c) At the particle is at rest at t=0 x=0. At this instant a driving force is switched on F = F_{0}cos(\omega*t) what is the equation of motion for t > 0 ? [2]
\ddot{x}= (F_{0}/m)cos(\omega*t) -(b/m)v
(d) Show that, when both forces are present, x(t)= A*cos(ωt−δ) is a solution to the equation of motion with appropriate choice of A and δ. Find A and δ .
We are also told that [cos(δ) = 1/(rootof 1 +tan^2(δ)] and sin(δ) = tan(δ)/(rootof 1 +tan^2(δ)]
Again I try to differentiate through but I come to a block, I use the 2 above identities but doesn't help me? I know I must be missing something simple or just not seeing it.
Any help much appreciated.
Homework Statement
A particle moving along the x-axis with velocity v experiences a resistive force –bv, but no spring-like restoring force, where b is the friction coefficient.
(a) Write down the equation of motion. [3]
\ddot{x}= -(b/m)v
(b) Show that the equation of motion is satisfied by
x(t) = C - (v_{0}/\gamma)e^{-\gamma}
where , m is the mass of the particle, and C and \gamma=(b/m) are free parameters. [3]
Not the quickest with latex so the answer was to just differentiate twice, you can then see they are equivalent.
(c) At the particle is at rest at t=0 x=0. At this instant a driving force is switched on F = F_{0}cos(\omega*t) what is the equation of motion for t > 0 ? [2]
\ddot{x}= (F_{0}/m)cos(\omega*t) -(b/m)v
(d) Show that, when both forces are present, x(t)= A*cos(ωt−δ) is a solution to the equation of motion with appropriate choice of A and δ. Find A and δ .
We are also told that [cos(δ) = 1/(rootof 1 +tan^2(δ)] and sin(δ) = tan(δ)/(rootof 1 +tan^2(δ)]
Again I try to differentiate through but I come to a block, I use the 2 above identities but doesn't help me? I know I must be missing something simple or just not seeing it.
Any help much appreciated.