so there are two forces acting on the P, a resorting force and an additional force that is forcing the system to oscillate.Quote:

http://img152.imageshack.us/img152/348/picture2os4.png

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 $\displaystyle \frac{1}{4} \sin 3t$ meters.

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

a) show that $\displaystyle \ddot{x} + 4 x = \sin 3t$

b) Find the first time after the motion starts, at which P is instantaneously at rest.

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

we are giving that the displace $\displaystyle s$ due to the addition force is $\displaystyle \frac{1}{4} \sin 3t$ so the acceleration due to this force at time $\displaystyle t$ is $\displaystyle -\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 $\displaystyle m \ddot{x} = -\frac{2 x }{0.5} + m \frac{9}{4} \sin 3t $ by Newton's Second Law

$\displaystyle \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$\displaystyle A\cos 2t+ B \sin 2t$

and the particular integral as $\displaystyle - \frac{9}{20}\sin 3t$

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

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

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

$

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

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

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

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

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

Many Thanks

Bobak