M3: S.H.M

for part c,

I calculated the period:

2\pi /\sqrt{40}

the amount of time spent within 0.25m of equilibrium of each spring:

using the below equation gives me the time from the eq position to 0.25 m - I doubled it because it will pass through that section twice in one complete period.

from eq to b:

0.25 = (1.8)sin((\sqrt{40})(t)) => t = 0.022... => 2t = 0.044...

from eq to a:

0.25 = (3.2)sin((\sqrt{40})(t)) => t = 0.078... => 2t = 0.156......

proportion: (total time spend within 0.25m of eq)/period

????

where have I gone wrong?

(edited 6 years ago)

Screen Shot 2018-04-05 at 13.22.32.png325.7KB

bump

in the mark scheme, they have a different a.

why are we using the time for a complete oscillation when the particle doesn't complete a complete oscillation.

how does this equal amplitude (a):

v/w = a

???? theyre a = 1/2 for both sides of the equation, why is that?

Screen Shot 2018-04-05 at 14.02.25.png59.2KB

Reply 3

Shaanv

V^2=w^2(a^2-x^2)
Max velocity when x=0
V^2=w^2*a^2
V=wa

The particle is given a speed of root(10) it is at equilibrium posioton, when x=0

(edited 6 years ago)

Reply 4

Maths&physics

Original post by Shaanv

V^2=w^2(a^2-x^2)
Velocity is at a maximum when x=0.
V^2=(w^2)*(a^2)
V=wa.

What makes u think that the particle doesnt undergo complete oscillations?

it says it doesn't before part b.

Reply 5

Maths&physics

Original post by Maths&physics

it says it doesn't before part b.

so, why is the amplitude for both sides? wouldn't the resistance in each string be different?

Reply 6

Shaanv

Original post by Maths&amp

it says it doesn't before part b.

It doesnt say this. It says that it doesnt reach A or B. There is no mention of whether or not P completes full oscillations.

Reply 7

Maths&physics

Original post by Shaanv

It doesnt say this. It says that it doesnt reach A or B. There is no mention of whether or not P completes full oscillations.

I thought a full oscillation was from A to B. is it because its a string and only the extension is considered a part of the oscillation? and due to the strings being different, wouldn't there be different resistance to motion in each string?

Reply 8

Shaanv

Original post by Maths&amp

They are not strings they are springs.

U have proved that the system undergoes SHM. U have got a value for w^2 for the system. So u can work out time period for one oscillation, and the amplitude if given a speed as in this case.

The amplitude will be the same either side of the equilibrium position.

I dont know why u are talking about resistance in each string.

There is tension in both springs, if particle is displaced from equilibrium the tensions in the two springs will change, so the resultant force acts towards equilibrium position.

Reply 9

Maths&physics

Original post by Shaanv

thanks

(edited 6 years ago)

Reply 10

Shaanv

Original post by Maths&amp

thanks

This graph represents the motion over one time period.

The question is asking you to work out the proportion of time that the the particle is within 0.25m of equilibrium. So the time the particle is between 0.25m and -0.25m.

The time ‘t’ given in the markscheme is the time it takes to travel from the equilibrium position to 0.25m.

From the diagram you can see there is 4 times in the particles motion that the particle is within 0.25m of the equilibrium position.

Due to the symmetry of the curve each of these 4 periods in time must be the same.

Thus total time that the particle is within 0.25m of equilibrium is ‘4t’.

The total time for one oscillation is the time period ‘T’, so 4t/T is the proportion of time in each oscillation that the particle is within 0.25m of equilibrium position.

Reply 11

Maths&physics

Original post by Shaanv

yeah, thanks a lot but I figured that out. when I was using the wrong amp, I realised that the equation was going to give me the time from the eq to 0.25, but the particle will past by there again (traveling at the same speed, same distance but in the opposite direction), which would mean I just need to double it. and once it goes past the eq and travels an equal but negative distance, it will be traveling at the same speed (although a negative velocity) and so it will be the same as the opposite side. therefore you just multiply the original time by 4. por yes, you can use the graph to prove there will be 4 times. either way, thanks a lot.