M3 Simple Harmonic Motion help!

I am struggling with part c of this question, and how to find the total time.

A particle P moves on the x-axis with simple harmonic motion about the origin O as centre. When P is a distance 0.04 m from O, its speed is 0.2 m/s and the magnitude of its acceleration is 1 ms–2.

a) Find the period of the motion.

T=2pi/5 using a = -(w^2)x

The amplitude of the motion is a metres.
Find

(b) the value of a,

using v^2=w^2(a^2-x^2) I got a = 0.057

(c) the total time, within one complete oscillation, for which the distance OP is greater than 0.5a metres.

For this, I have used x=asin(wt) with x = 0.5a, and achieved t = pi/30.
I used the sine graph to calculate the 2 places where distance OP is greater than 0.5a metres, and got pi/15 seconds as final anwer.

But in the mark scheme it has the total time as T = 4pi/15

Can someone please tell me how I should go about solving this?! Thank you very much!! :smile:

Reply 1

pleasedtobeatyou

Original post by Coral Reafs

...

As you rightly deduced

T= \frac{2 \pi}{5}

Consider just the basic version of the sine curve.

y = sin(x)

The maximum "amplitude" is 1. The locations where the amplitude is 0.5 are

\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6} and \frac{11 \pi}{6}

. The total "time" between each the first/second and third/fourth points is

\frac{4 \pi}{6}

. Multiply this by 2 because there are two sets of these points -->

\frac{8 \pi}{6} = \frac{4 \pi}{3}

Now, we divide this by

2 \pi

to give the proportion of the period which is greater than half of the total amplitude. In this example

T_{total}

has a period of 1.

T_{over 0.5a} = \frac{2}{3} \cdot T_{total}

So, in the case of the question...

T_{over 0.5a} = \frac{2}{3} \cdot T = \frac{2}{3} \cdot \frac{2 \pi}{5} = \frac{4 \pi}{15}

(edited 9 years ago)

Reply 2

Coral Reafs

Original post by pleasedtobeatyou

As you rightly deduced

T= \frac{2 \pi}{5}

Consider just the basic version of the sine curve.

y = sin(x)

The maximum "amplitude" is 1. The locations where the amplitude is 0.5 are

\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{7 \pi}{6} and \frac{11 \pi}{6}

. The total "time" between each the first/second and third/fourth points is

\frac{4 \pi}{6}

. Multiply this by 2 because there are two sets of these points -->

\frac{8 \pi}{6} = \frac{4 \pi}{3}

Now, we divide this by

2 \pi

to give the proportion of the period which is greater than half of the total amplitude. In this example

T_{total}

has a period of 1.

T_{over 0.5a} = \frac{2}{3} \cdot T_{total}

So, in the case of the question...

T_{over 0.5a} = \frac{2}{3} \cdot T = \frac{2}{3} \cdot \frac{2 \pi}{5} = \frac{4 \pi}{15}

Alright, so I understand the period is 1 due to one full oscillation, and according to the sine curve, as you mentioned pi/6, and 5pi/6 are the places where distance is more than 0.5a, but you also mentioned 7pi/6 and 11pi/6, and these points on the graph are negative values, however they are more than -0.5a, so do I include the negative values of a which are below -0.5a and include the times in my answer too?

By the way, thank you so much for your time and effort! +1 :smile:

Reply 3

Dilzo999

Firstly always on these time questions I start off and draw a diagram, it literally reveals all the answers and can help you when you're confused. I've attached an image with my working out and tips :wink:

Reply 4

pleasedtobeatyou

Original post by Coral Reafs

Yes because the third part of your question asks for the times when the distance is greater that 0.5a

Distances are always positive, so this implies they are looking for the times where the magnitude of the displacement is greater than 0.5a

Hence, these two times correspond the the two regions on the sine curve that Dilzo999 has drawn.

Reply 5

Coral Reafs

Original post by Dilzo999

Thank you so so much for your effort into the answer, it is much appreciated, and I got the same graph as you did, it's just I didn't know how to interpret it before, so thank you! :smile:

Reply 6

Coral Reafs

Original post by pleasedtobeatyou

...

Original post by Dilzo999

...

Btw sorry to bother you guys again, but is there any specific scenarios as into when you would use x=acoswt and x=asinwt? because it seems whenever using x=asinwt, there is always more working involved rather than using x=acoswt in the markschemes.

Thanks alot!