# Hw 11 Problem 5

3,669 views

A particle moves along the x axis. It is initially at the position 0.230 m, moving with velocity 0.220 m/s and acceleration -0.210 m/s2. Suppose it moves with constant acceleration for 3.20 s.

(a) Find the position of the particle after this time.
()  m
I found this to be -0.1412
(b) Find its velocity at the end of this time interval.
() m/s

I got -0.452 for part b, but I have no idea how to go about the rest. I tried the practice problems several times and couldn't get any method to work.

We take the same particle and give it the same initial conditions as before. Instead of having a constant acceleration, it oscillates in simple harmonic motion for 3.20 s around the equilibrium position x = 0.

(c) Find the angular frequency of the oscillation. Hint: in SHM, a is proportional to x.
()  /s

(d) Find the amplitude of the oscillation. Hint: use conservation of energy.
()  m

(e) Find its phase constant 0 if cosine is used for the equation of motion. Hint: when taking the inverse of a trig function, there are always two angles but your calculator will tell you only one and you must decide which of the two angles you need.

(f) Find its position after it oscillates for 3.20 s.
()  m

(g) Find its velocity at the end of this 3.20 s time interval.
()  m/s

For simple harmonic oscillation, you have $$x(t)=Acos(\omega t+\phi _0)$$ $$v(t)=-\omega Asin(\omega t+\phi _0)$$ $$a(t)=-\omega ^2Acos(\omega t+\phi _0)$$

Set $t=0$ as the initial state and we have: $$x(0)=Acos(\omega 0+\phi _0)$$ $$v(0)=-\omega Asin(\omega 0+\phi _0)$$ $$a(0)=-\omega ^2Acos(\omega 0+\phi _0)$$

To find $\omega$, divide the third equation by the first equation and you have $$a=-\omega ^2x$$

Plug in the numbers and you find $\omega$.

In general, simple harmonic oscillation requires that $F_{net}=ma=-constant\cdot x$. In terms of $\omega$, we can write: $$F=-m\omega ^2x$$

This holds to all simple harmonic oscillations including the mass-spring cases. We can consider an effective spring constant $k_e$ for the general case and we would have $F=-k_ex$. $$\therefore k_e=m\omega ^2$$.

With $k_e$, one can calculate the potential energy as $U=\frac 1 2 k_ex^2$.

Then the total energy can be written as $$E=K+U=\frac 1 2 k_eA^2=\frac 1 2mv^2+\frac 1 2k_ex^2$$

Substitute with $k_e=m\omega ^2$ and you have $$\frac 1 2 m\omega ^2A^2=\frac 1 2mv^2+\frac 1 2m\omega ^2x^2$$

Plug in values at $t=0$, one can find A.

The rest part can be solved by plug in numbers in the set of motion equations listed above.
answered Oct 25, 2013 by (21,750 points)
But what does m equal to find k(sub)e?