This discussion is now closed.

Check out other Related discussions

M3 further dynamics SHM

A bouy on the surface of te sea is tied to a pier and oscillates in a vertical line with SHM .the period of the oscillation is 3pi/4 s, and the maximum speed of the bouy is 4ms^-1.At the highest point of the motion the bouy is 5m below the pier .Calculate the amplitude of the motion and find the time taken by the bouy to fall from its highest point to 6m below the pier...

I am a bit confused ont the 2nd bit ...The book takes at t=0 the highest point which makes it a cosine graph(acoswt)..
Is there any reason how they came to choose that it was at the highest point when t=0 or did they choose it arbitrary..

Reply 1

Trevor 12345

3

rbnphlp

A bouy on the surface of te sea is tied to a pier and oscillates in a vertical line with SHM .the period of the oscillation is 3pi/4 s, and the maximum speed of the bouy is 4ms^-1.At the highest point of the motion the bouy is 5m below the pier .Calculate the amplitude of the motion and find the time taken by the bouy to fall from its highest point to 6m below the pier...

I am a bit confused ont the 2nd bit ...The book takes at t=0 the highest point which makes it a cosine graph(acoswt)..
Is there any reason how they came to choose that it was at the highest point when t=0 or did they choose it arbitrary..

It makes no difference whatsoever. It has been chosen for now other reason than mere convenience.

Reply 2

TheNack

From what I remember of my A-level SHM stuff, it's pretty much an arbitrary decision.

Reply 3

tiny hobbit

15

Since you're asked to find the time from the highest point, it seems reasonable to take that position as t=0. You then have to use x=acoswt since x is measured from the centre of the motion and putting x=0 into x=acoswt gives x=a. Now put in x = a-1 (I take it you've already worked out a) then solve the eqaution to find t.

If you wanted to measure time from the centre of the motion you would use x=asinwt.

Reply 4

james.h

11

rbnphlp

A bouy on the surface of te sea is tied to a pier and oscillates in a vertical line with SHM .the period of the oscillation is 3pi/4 s, and the maximum speed of the bouy is 4ms^-1.At the highest point of the motion the bouy is 5m below the pier .Calculate the amplitude of the motion and find the time taken by the bouy to fall from its highest point to 6m below the pier...

I am a bit confused ont the 2nd bit ...The book takes at t=0 the highest point which makes it a cosine graph(acoswt)..
Is there any reason how they came to choose that it was at the highest point when t=0 or did they choose it arbitrary..

Time itself is continuous, so the choice of an artificial zero-value for time is arbitrary. This means you can choose any point in its motion as being at t=0, and from that you might as well choose t=0 at the start of the interval you're interested in. :yep:

Just makes things easier.

Reply 5

rbnphlp

OP

11

Thanks everyone got it ..

Reply 6

rbnphlp

OP

11

The water level in a harbour can be assumed to rise and fall with SHM.On a certain day low tide,will occur at 12 noon and the depth of the water will be 3m.The subsequent high tide will occur at 6.20pm and the water will be 15m.A ship which needs 6m depth of water wishes to use harbour.Find to the nearest minute, the earliest it can enter the harbour on this day and the time by which it must leave.

Am I right in thinking the amplitude is 6m....
and x=asinwt is the earliest time where x=a , How do I find T?
Could some one point me in the right direction...

OP

anyone?

rbnphlp

The water level in a harbour can be assumed to rise and fall with SHM.On a certain day low tide,will occur at 12 noon and the depth of the water will be 3m.The subsequent high tide will occur at 6.20 and the water will be 15m.A ship which needs 6m depth of water wishes to use harbour.Find to the nearest minute, the earliest it can enter the harbour on this day and the time by which it must leave.

Am I right in thinking the amplitude is 6m....
and x=asinwt is the earliest time where x=a , How do I find T?
Could some one point me in the right direction...

T is the period of oscillation, ie: the time that elapses between two consecutive peaks. As this motion is modelled by SHM, this means that the period can be measured as 2 times the time between high tide and low tide (symmetrical wave). Then omega is 2 pi over T, which you knew already. :smile:

I think that's right. :yep:

Reply 9

rbnphlp

OP

11

james.h

T is the period of oscillation, ie: the time that elapses between two consecutive peaks. As this motion is modelled by SHM, this means that the period can be measured as 2 times the time between high tide and low tide (symmetrical wave). Then omega is 2 pi over T, which you knew already. :smile:

I think that's right. :yep:

ok so t should be just t=760min, but T should be in secs for the equation to work ..but the question say they want the time T to the nearest minute?

Reply 10

rbnphlp

OP

11

A small doubt ..It would be great if someone could clear this up for me..

A particle moving in SHM has a =5m, find the time when x=3,
say w=2
when I use x=asinwt or x=acoswt(I know you choose according to your convenience)..
Could some one say what x is in both cases in (asinwt and coswt)and why?
Thanks and will rep+

OP

please anyone :sad:

Where did it start (or alternatively do you know a time when it is at a particular place)?

The t in either equation is effectively a length of time, so if we don't know when/where we are measuring time from, we don't know what sum to do.

Reply 13

rbnphlp

OP

11

tiny hobbit

Where did it start (or alternatively do you know a time when it is at a particular place)?

The t in either equation is effectively a length of time, so if we don't know when/where we are measuring time from, we don't know what sum to do.

Dont worry I sorted it out ..thanks anyway.. :smile:

M3 further dynamics SHM

Related discussions

Latest

Trending

Trending

Articles for you