1. ## Forced Harmonic Motion

The figure shows a particle P of mass 1kg which is free to slide horizontally inside a smooth cylindrical tube. The particle is attached to one end of a light elastic spring of natural length 0.5 m and modulus of elasticity 2 N.

The system is initially at rest. The other end Q of the spring is then forced to oscillate with simple harmonic motion so that at time t second its displacement from its initial position is $\frac{1}{4} \sin 3t$ meters.

The displacement of P from its initial positions at time $t$ seconds is $x$ meters, measured in the same direction as the displacement of Q

a) show that $\ddot{x} + 4 x = \sin 3t$
b) Find the first time after the motion starts, at which P is instantaneously at rest.
so there are two forces acting on the P, a resorting force and an additional force that is forcing the system to oscillate.

the resorting force is due to the tension in the spring. $T = \frac{\lambda x }{l}$ by hookes law.

we are giving that the displace $s$ due to the addition force is $\frac{1}{4} \sin 3t$ so the acceleration due to this force at time $t$ is $-\frac{9}{4} \sin 3t$ (by taking the derivative with respect to time twice)

both the resorting force and the additional force are opposing the direction of motion.

so $m \ddot{x} = -\frac{2 x }{0.5} + m \frac{9}{4} \sin 3t$ by Newton's Second Law

$\Rightarrow \ddot{x} +4x = \frac{9}{4} \sin 3t$ which disagrees with the questions, who right me or them ???

For the next part I solved the differential equation.

I get the complementary function as $A\cos 2t+ B \sin 2t$
and the particular integral as $- \frac{9}{20}\sin 3t$

so I have $x= A\cos 2t+ B \sin 2t - \frac{9}{20}\sin 3t$

A is obviously zero as the particle had zero displacement initially and we are given that the derivative is zero at t = 0

$\dot{x} = 2B \cos 2t - \frac{27}{20}\cos 3t
$

so $A = 0, B = \frac{27}{40}$

so i must solve the equation $\frac{27}{20}\cos 2t - \frac{27}{20}\cos 3t = 0$ so find out when this particle is at rest.

some simple trig gives $\frac{27}{20} \left( \sin\left( \frac{5}{2}t \right)\sin\left( \frac{1}{2}t \right)\right) = 0$

$\frac{5}{2}t = n \pi$ or $\frac{1}{2}t = n \pi$

the smallest solution is given by $t = \frac{2 \pi}{5}$. which agress with the book. So is my method good, or not ?

Many Thanks

Bobak

2. Originally Posted by bobak
so there are two forces acting on the P, a resorting force and an additional force that is forcing the system to oscillate. There is only one force acting on the particle: the tension (or compression) in the spring.

the resorting force is due to the tension in the spring. $T = \frac{\lambda x }{l}$ by hookes law.

we are giving that the displace $s$ due to the addition force is $\frac{1}{4} \sin 3t$ so the acceleration due to this force at time $t$ is $-\frac{9}{4} \sin 3t$ (by taking the derivative with respect to time twice) Sorry, but that is nonsense. You can't just differentiate what's happening at one end of the spring in prder to find out what's happening at the other end.
At time t, the length of the spring is $x(t) - {\textstyle\frac14}\sin3t$. So the force on the particle is $\frac{2(x(t) - {\textstyle\frac14}\sin3t)}{0.5}$ (Hooke's law).

That gives the equation of motion as in the book: $\ddot{x} = -4(x - {\textstyle\frac14}\sin3t)$, by Newton's second law.

The solution that fits the initial conditions is $\textstyle x(t) = \frac3{10}\sin2t - \frac15\sin3t$, and the particle is at rest when $\textstyle \frac35(\cos2t - \cos3t)=0$. This is the same equation that you arrived at, apart from the numerical multiple 3/5 instead of 27/20. So it gives the same solution.

3. Originally Posted by Opalg
Sorry, but that is nonsense. You can't just differentiate what's happening at one end of the spring in prder to find out what's happening at the other end.
Thanks for that, I realise how little sense that makes now.

Best Regards

Bobak