Applications of Calculus to the Physical World (3 Viewers)

[FDownes]

I've got a whole pile of questions I need help with, but I have restricted access to the internet at the moment, so I mightn't be able to check back here for a little while. Lets hope you guys can help me out.

1. If a the displacement of a particle moving in simple harmonic motion is given by x = 5cos3t + 2sin3t, find the maximum speed of the particle.
2. A particle is moving in simple harmonic motion with acceleration d²x/dt² = -4xms^-1. If the particle starts at the origin with a velocity of 3ms^-1, find; a) The endpoints of its motion, b) The exact speed when the particle is 1m from the origin.
3. A weight is suspended from a spring and pulled down to its maximum displacement of 9cm, and then let go. Its acceleration is given by a = -1/9cm^-2. Find; a) The equation of its velocity in terms of x, b) Its exact position when its velocity is 2cms^-1
4. The period of a particle moving in SHM is 8cm. Calculate its velocity and acceleration (correct to 1 decimal place) when the displacement is 5cm from the center of motion.
5. A particle moves in a straight line so that its acceleration at any time is given by d²x/dt² = -9x. Find its period, amplitude and displacement at time t if initially the particle is 2cm from the origin and has velocity √3cm^-1.
6. A particle moves in a line so that its acceleration is given by d²x/dt² = 8 - 2x. Initially, the particle is at the origin and has velocity 3√2ms^-1; a) Find the interval in which the particle will travel, b) Is the motion of the particle in simple harmonic motion?
   [*]A weight is oscillating at the end of a spring, with velocity given by v² = 900 - 1600x²cmw^-1; a) Find the acceleration of the weight with respect to x, b) Find the period of the motion, c) Find the maximum velocity of the weight.
   [*]a) If x = a sin nt + b cos nt, find the acceleration of the particle in terms of t, and show that acceleration = -n²x, b) Find the amplitude and period of the motion, c) Find the maximum velocity.
   [*]A particle moves in simple harmonic motion with amplitude 5cm and period 6 seconds. Find; a) The velocity when the particle is 2.5cm from the center of motion, b) The maximum acceleration.
   [*]A missile is launched at an initial trajectory of 68^o and a velocity of 1200ms^-1. Neglecting air resistance anf the curvature of the earth and taking the acceleration due to gravity as 9.8ms^-2, calculate; a) The time taken for its flight (to the nearest minute), b) How far away it will hit its target (to the nearest km).
   [*]An arrow is fired at a velocity of 24ms^-1 and is aimed at the center of a target 1m high and 35m away. The air resitance is negligable for angles of projection less than 45^o. At what angle should the arrow be fired? Use g = 10ms^-2.
   [*]A gun is aimed at a target on the ground 150m away. If the initial velocity is 125ms^-1, find the angles at which the gun could be fired to reach the target (use g = 10ms^-2).

I know this is a lot to ask, so I wouldn't expect one person to go through and answer every question (although that would be cool). But if everyone can work together to wade through the torrent of questions it'd be greatly appreciated.

 

[vds700]

ill start with Q1

x = 2sin3t + 5cos3t = Rsin(3t + A)
2sin3t + 5cos3t = Rsin3tcosA + Rcos3tsinA
equate coefficents of sin and cos

RcosA = 2 ----(1)
RsinA = 5 ----(2)

square and add 1 and 2
R^2 = 4 + 25 = 29
R = sqrt29

2/1

RsinA/RcosA = 5/2
tanA = 5/2
A = arctan(5/2)

therefore x = sqrt29sin(3t + arctan(5/2))
v = dx/dt = 3sqrt29cos(st + arctan(5/2))

the amplitude of the velocity vs time graph is 3sqrt29, which is the max speed.


R

 

[FDownes]

Thanks, we're off to a good start.

 

[anna-goanna123]

For question 2, is it meant to be d2x/dt2 (sorry don't know how to write powers on the computer.. i am hopeless!) = -4x or =-4t??? because if it has been differentiated with respect to t, then it should be =-4t. If so, then i know how to answer it, but if it is actuall =-4x, then im stumped lol. fat load of help i am. I dunno if i'm making any sense lol!

 

[donetha]

For question 2, is it meant to be d2x/dt2 (sorry don't know how to write powers on the computer.. i am hopeless!) = -4x or =-4t??? because if it has been differentiated with respect to t, then it should be =-4t. If so, then i know how to answer it, but if it is actuall =-4x, then im stumped lol. fat load of help i am. I dunno if i'm making any sense lol!

Its supposed to be -4x...the proof of SHM. That x:=-n^2x. In this case n=2

 

[vds700]

For question 2, is it meant to be d2x/dt2 (sorry don't know how to write powers on the computer.. i am hopeless!) = -4x or =-4t??? because if it has been differentiated with respect to t, then it should be =-4t. If so, then i know how to answer it, but if it is actuall =-4x, then im stumped lol. fat load of help i am. I dunno if i'm making any sense lol!

it just means a = -4x. So n = 2

then using v² =n²(a² - x²)
when x = 0, v = 3
9 = 4a²
a² = 9/4
a = 3/2

 

[anna-goanna123]

For question 2 then, is this right anybody, cause i need this practise lol???

find velocity
where v= dx/dt= -2t^2 +c
but at t=0, v=3, so
c=3

the endpoints are where v=0
ie -2t^2 + 3 = 0
(you can solve this cause im hopeless at factorising!)

and part b) x=-2t^2 +3 +k
where t=0, x=0
therefore k=-3
so x= -2\3t^3 +3t - 3!
sub in x=1
ie 1= -2/3^3 +3t -3
and again u can solve from here

is this right anyone, or have i gone hopelessly in the wrong direction? ive got an exam on this stuff coming up next week and i can't remember anything lol!

 

[vds700]

For question 2 then, is this right anybody, cause i need this practise lol???

find velocity
where v= dx/dt= -2t^2 +c
but at t=0, v=3, so
c=3

the endpoints are where v=0
ie -2t^2 + 3 = 0
(you can solve this cause im hopeless at factorising!)

and part b) x=-2t^2 +3 +k
where t=0, x=0
therefore k=-3
so x= -2\3t^3 +3t - 3!
sub in x=1
ie 1= -2/3^3 +3t -3
and again u can solve from here

is this right anyone, or have i gone hopelessly in the wrong direction? ive got an exam on this stuff coming up next week and i can't remember anything lol!

the acceleration is -4x not -4t. I think u may be confused by the way they word the question, they say d²x/dt² = -4x. Let me explain, there are a number of forms for acceleration that can be used, dv/dt, d²x/dt², v. dv/dx, d/dx(1/2 v²) etc. They mean the same thing, so if u needed to find the velocity, u would say:

d/dx(1/2 v²) = -4s then integrate with respect to x

But this is not really necessary because u can solve it the way i did using the fformula v² =n²(a² - x²). a = 3/2 so the endpoints are +/- 3/2 metres. Then to find the speed when x = 1, sub in x = 1 into the formula.

Hope this helped annagoanna and Fdownes

 

[FDownes]

Thanks vds700 for clearing that up.

 

[cyrius]

For question 3. Is it a= -1/9cms^-2?
Anyways given is a= -1/9cms^-2 and when v=0 x=9 (not sure with this)
d/dx(1/2v²)=∫-1/9 dx
a) 1/2v²=-1/9x + C
v²=-2/9x + 2C
v= √-2/9x + 2C

but when v=0 x=9
0²=-2/9(9) + 2C
2C = 2
v= √-2/9x + 2 cms<sup>-1

</sup> b) when v=2
v= √-2/9x + 2
(2= √-2/9x + 2)²
[FONT=&quot]4=-2/9x + 2
9(2=-2/9x)9
18=-2x
x = -9 cm

I hope this is right...LOL
[/FONT]


 

[davidgoes4wce]

Seeing that no one had an attempt at Q7. (Q18 E6.10 of Grove text) I have had a go at that question (this is my first attempt at answering any questions on the HSC Board of Studies forum)

7a)
v^2=900-1600x^2

v^2=2(450-800x^2)
v^2 / 2 =450-800x^2

d/dx (v^2 / 2) = -1600 x

Answer = -1600 x

b)
a=-n^2 x = - (40^2) x
n= 40

T=2 pi/ n
= 2 pi/ 40
= pi / 20

Answer= pi/20

c) v^2=n^2 (a^2 -x^2)
let x =0
v^2= 900
v=-30 or 30

Answer: v=-30 or v=30

 

[Speed6]

(Spongebob Squarepants voice)...sevun years laytur

 

[enigma_1]

Seeing that no one had an attempt at Q7. (Q18 E6.10 of Grove text) I have had a go at that question (this is my first attempt at answering any questions on the HSC Board of Studies forum)

7a)
v^2=900-1600x^2

v^2=2(450-800x^2)
v^2 / 2 =450-800x^2

d/dx (v^2 / 2) = -1600 x

Answer = -1600 x

b)
a=-n^2 x = - (40^2) x
n= 40

T=2 pi/ n
= 2 pi/ 40
= pi / 20

Answer= pi/20

c) v^2=n^2 (a^2 -x^2)
let x =0
v^2= 900
v=-30 or 30

Answer: v=-30 or v=30

y u fkn bump a thread with mif questions

 

[Speed6]

y u fkn bump a thread with mif questions

Ikr, just provided Margaret Grove with some more ego to publish her next fkn next textbook. F**K!!!!

 

[InteGrand]

Would anyone be complaining if it weren't known where these questions were from?

 

[davidgoes4wce]

I also had an attempt at FDownes Q8 (From Grove E6.10 Q19)

a) if x=a sin nt + b cos nt, find the acceleration of the particle in terms of t, and show that a=-n^2 x

x= a sin nt + b cos nt
v= ab cos nt - bn sin nt
a= -a n^2 sin nt - bn^2 cos nt
=-n^2 x (factorising)

b) Find the amplitude and period of the motion

Using the Auxillary Angle Method

a sin B + b cos B = A sin (B + C)
a sin nt + b cos nt= A sin (B + C)

Our task is to solve for A,B and C

1. Solving for 'A'
where A= sqrt(a^2 + b^2)

2. Solving for 'B'
B=nt

3. Solving for 'C'
tan C=b/a
tan C=nt/nt
tan C=1
C= pi/4

The auxillary equation is:

x= A sin (nt + pi/4)
'A' represents the Amplitude which is sqrt(a^2 + b^2)
Period = 2pi/n

c)
v=dx/dt
= An cos (nt + pi/4)

Amplitude of velocity is therefore 'An'

=An
=sqrt(a^2+b^2) n

 

[davidgoes4wce]

y u fkn bump a thread with mif questions

I am going through those questions because I have nothing to do on a Saturday night.

 

[Speed6]

I am going through those questions because I have nothing to do on a Saturday night.

You do know the rep of mif in the society of BOS right?

 

[davidgoes4wce]

You do know the rep of mif in the society of BOS right?

No I don't please inform me.

 

[davidgoes4wce]

I use all questions from : Fitzpatrick, Grove, Patel, Cambridge, past papers. So I will try to answer them as best as I can. I do use these forums a lot to help me with my own work. With so many questions, just want to improve my knowledge of this subject.