# Simple harmonic motion frequency [closed]

I already did part a and b by using kinematics but I'm stuck for the next part

A particle moves along the x axis. It is initially at the position 0.300 m, moving with velocity $0.070 m/s$ and acceleration $-0.330 m/s^2$. Suppose it moves with constant acceleration for $5.60 s$.

• a. Find the position of the particle after this time.
• b. Find its velocity at the end of this time interval.

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 $5.60 s$ around the equilibrium position $x = 0$.

• c. Find the angular frequency of the oscillation. Hint: in SHM, $a$ is proportional to $x$.

This is what I tried and I am within 10 %. I kept all the decimals so there is probably something that I am missing

$$f = \frac{1}{T} = \frac{1}{2 \pi}\sqrt\frac{k}{m}$$

$$\omega = 2\pi f$$

The given period is $5.60$ sec (I believe so but the wording is not completely clear).

The frequency is then:

$$f = \frac{1}{T} = .1785714 \: Hz$$

$$\omega = 2 \pi f = 1.1219973 \: rads/sec$$ and i get within 10%

## closed as off-topic by John Rennie, Jim, David Z♦Nov 11 '14 at 16:36

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• I think this is fine as a homework question as the OP has made a sensible attempt at the question and wants to know why it is wrong. – nivag Nov 11 '14 at 16:47

I think you are misinterpreting the question, although I agree its not that clearly written. I do not think time is the period of the motion (is there a part d?).

I would use the initial position, velocity and acceleration to work out $\omega$. (Use the hint in the question).

• why did the solution Autolatry suggested work. they give us constant acceleration of -.33 for parts and b. for part c we don't know if the object in fact has an acceleration of -.33 at position .3 how come is it correct? – Petrarca Nov 11 '14 at 16:07
• @Petrarca Yes you do. It says it is the same particle at the same initial conditions. You have to read the question very carefully. – nivag Nov 11 '14 at 16:15

In SHM, the acceleration of a moving body is related to it's displacement by; $$a(x) = - \omega^{2} x$$ It's easy to see from your question, that we have $$-0.33 = -0.3\omega^{2}$$ From which we obtain \begin{eqnarray} \omega^{2} &=& \frac{0.33}{0.3} \\ &=& 1.1 \end{eqnarray} From which $$\omega = \sqrt{1.1} = 1.049 s^{-1}$$

• It's important to remember that I have clearly taken the positive sqare root in this example. – Autolatry Nov 11 '14 at 15:41
• For homework type questions you should try and avoid just giving the answer. Instead say where people are going wrong or suggest a suitable method and encourage them to work it out for themselves. See this meta question – nivag Nov 11 '14 at 15:46
• my problem with that approach is that they give us constant acceleration of -.33 for parts and b. for part c we don't know if the object in fact has an acceleration of -.33 at position .3 . am i missing something? (ALTHOUGH IT IS CORRECT BUT I DONT SEE HOW) – Petrarca Nov 11 '14 at 16:00
• The question states that intitially the body moves from an initial displacement of $o.3$m for $5.6$s with constant acceleration. – Autolatry Nov 11 '14 at 16:19