Simple Harmonic Motion HELP! (1 Viewer)

[pomsky]

YO YO YO
Can you guys help me with the below Qs? My brain cannot cope with physics and it knows it (and yet this is math ext 1 wtf). It would be great if you guys could provide some working out and/or give me tips on how to approach these questions.thhhaankkks

Oh and hi everybody! I'm a n00b here- looking forward to being bored of studies with you all (or are we already? haha)

1. A particle moves according to x=3-2cos^2(2t) in units of centimetres and seconds.
a.) Express the equation in the form x=x0-Acos(wt)
Ans: x=2-cos(4t)

b.) Find the range of motion.
Ans: 1<=x<=3

c.) What is the maximum speed of the particle and when does it first occur?
Ans:4cm/s; pi/8 seconds

2. A particle P moves so that its acceleration is proportional to its displacement x from a fixed point O and opposite in direction. Initially the particle is at the origin, moving with velocity 12m/s and the particle is stationary when x=4.

a.) Find a and v^2 as functions of x.
Ans: a=-9x,c^2=9(16-x^2)

b.) Find x,v and a as functions of t.
Ans: x=4sin3t, v=12cos3t, a=-36sin3t

c.) Find the displacement, acceleration and times when the particle is at rest.
Ans: x= +-4, a= +-36, t= (2k+1)*pi/6

d.) Find the velocity, acceleration and times when displacement is zero.
Ans: v=+-12, a=0, t=k*pi/3

 

[InteGrand]

YO YO YO
Can you guys help me with the below Qs? My brain cannot cope with physics and it knows it (and yet this is math ext 1 wtf). It would be great if you guys could provide some working out and/or give me tips on how to approach these questions.thhhaankkks

Oh and hi everybody! I'm a n00b here- looking forward to being bored of studies with you all (or are we already? haha)

1. A particle moves according to x=3-2cos^2(2t) in units of centimetres and seconds.
a.) Express the equation in the form x=x0-Acos(wt)
Ans: x=2-cos(4t)

b.) Find the range of motion.
Ans: 1<=x<=3

c.) What is the maximum speed of the particle and when does it first occur?
Ans:4cm/s; pi/8 seconds

2. A particle P moves so that its acceleration is proportional to its displacement x from a fixed point O and opposite in direction. Initially the particle is at the origin, moving with velocity 12m/s and the particle is stationary when x=4.

a.) Find a and v^2 as functions of x.
Ans: a=-9x,c^2=9(16-x^2)

b.) Find x,v and a as functions of t.
Ans: x=4sin3t, v=12cos3t, a=-36sin3t

c.) Find the displacement, acceleration and times when the particle is at rest.
Ans: x= +-4, a= +-36, t= (2k+1)*pi/6

d.) Find the velocity, acceleration and times when displacement is zero.
Ans: v=+-12, a=0, t=k*pi/3

1.
a) Use the trig. identity to make the squared cosine in the question into a cosine, then expand and simplify.

b) Once you have the equation for that's the answer to part a), note that a cosine function oscillates between +1 and -1. So the displacements extreme values are when that cosine is -1 or +1. When it is +1, the displacement is 2 - (+1) = 1, and when the cosine is -1, the displacement is 2 - (-1) = 3, hence x varies between 1 and 3.

c) Differentiate x(t)'s function from part a) to get the equation for v(t). This will be a sinusoidal function. You find it's extreme values in a similar way to how we found extreme values for displacement in part b). To find the first time this happens, you can set the function for velocity to 0, and solve, taking the smallest positive value for t as the answer.

2. The first sentence is the definition of S.H.M., so that tells us the particle moves in S.H.M.

The fact that the particle is stationary at x = 4 means that this is one of it's extreme values for displacement. This is because a particle undergoing S.H.M. stops when and only when it reaches its endpoints of the cycle (think of an object bobbing up and down on a spring, or visualize a pendulum). So the amplitude of this motion is 4, since the centre of motion is the origin (x = 0).

For a particle undergoing S.H.M. about the origin, the general equation for displacement is , where is what's called the angular frequency (it's equal to , where f is the frequency, which is how many cycles the particles completes per second), A is the amplitude (in this case equal to 4 as we've shown above), and is something called the phase shift or phase angle, which is determined based on where the particle started and its initial speed. Also, the S.H.M. differential equation for acceleration in terms of x is (this is just the fact that acceleration os negative proportional to displacement, and uses the fact that the proportionality constant is the square of the coefficient of t in the sine for the x(t) equation (usually you'll see instead of in textbooks.

To find , we can use the formula for S.H.M. .
This equation must hold at a times. We are told that at x = 0, we have v = 12. So plug in these values to that equation, remember that A = 4, and solve for . Then you will have completed part a).

For part b), since the particle starts at the origin, the phase angle must be 0 (you can see this by substituting x = 0 and t = 0 into , and then seeing that to have the RHS be 0, the sine must be 0, requiring the phase angle to be 0, as sin 0 = 0). Now that you'll have the equation for displacement as a function of time, you can differentiate to find similar ones for velocity and acceleration.

For part c), particle is at rest when the particle is at extreme points (imagine a pendulum), hence the given answer. Also, at 0, the acceleration is at extreme values, since acceleration is proportional to displacement, and displacement is at extrema. You find these extreme values for acceleration similar to how we did for displacement in Q.1. To find times when particle is at rest, set the function for velocity to 0, and solve using general solutions of trig. equations, and restrict the values of k so that t > 0.

For part d), use the equations from part a), and for the times when displacement is 0, set the equation for displacement as a function of time to 0, and solve using general solutions of trig. equations again.

 

[pomsky]

Wow thanks InteGrand!

Another quick question:
Assume that over several days of constant weather, the cycle of temperatures each day varies in simple harmonic motion between 13degrees celcius at 4am and 23 degrees celcius at 4pm. At what time(s), to the nearest minute, in the 12 hours from 4am to 4pm would the temperature be:
a.) 18OC
Ans:10am

b.) 15OC
Ans 7:33 am

c.) 21OC
Ans 12:27pm

thanks again

 

[pomsky]

bump~

 

[phoenix159]

Let the temperature (as a function of time) be T = a sin (nt + b) where t is the time in hours.

Set t = 0 as 4 AM and find the values of a, n and b.

Substitute the temperature values into the LHS of the equation and solve for t in each case.

 

[Librah]

Let the temperature (as a function of time) be T = a sin (nt + b) where t is the time in hours.

Set t = 0 as 4 AM and find the values of a, n and b.

Substitute the temperature values into the LHS of the equation and solve for t in each case.

You can make the graph alot nicer ie, in the form x=-5cos Zt/12 (Z is pi). By making 13 degrees -5 and 23 degrees as 5 on the y-axis with 18 degrees as your zero point, period is 24 hours so n=2pi/24. Then 0=-5cosnt/12 gives you t=6. Just add 6 hours to 4 am (0 point) and you have 10 am. Similarly 15 degrees will be x=-3 and 21 degrees is x=3.