Easy Simple Harmonic Motion Problem

In summary, at t=0 the lagging particle is just leaving one end of the path, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?At t=0, the lagging particle is just leaving one end of the path, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?The lagging particle
  • #1
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Homework Statement


Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.5 m. Each particle has a period of 1.5 s, but they differ in phase by π/5 rad. (a) How far apart are they 0.46 s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

Homework Equations


x=a*cos(wt+ø)
T=2π/w

The Attempt at a Solution


the amplitude is given by a=1.5m
and T=1.5s
then we have w=2π/T=4.18879..
the two particles are identical except they differ in phase by π/5
my attempt in finding the distance between them after t=0.46s is
d=1.5cos(wt)-1.5cos(wt+π/5)=0.28647
which does not match the correct solution
can someone point out where I went wrong
 
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  • #2
i_hate_math said:

Homework Statement


Two particles oscillate in simple harmonic motion along a common straight-line segment of length 1.5 m. Each particle has a period of 1.5 s, but they differ in phase by π/5 rad. (a) How far apart are they 0.46 s after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

Homework Equations


x=a*cos(wt+ø)
T=2π/w

The Attempt at a Solution


the amplitude is given by a=1.5m
and T=1.5s
then we have w=2π/T=4.18879..
the two particles are identical except they differ in phase by π/5
my attempt in finding the distance between them after t=0.46s is
d=1.5cos(wt)-1.5cos(wt+π/5)=0.28647
which does not match the correct solution
can someone point out where I went wrong
"How far apart are they 0.46 s after the lagging particle leaves one end of the path?"
At t=0, the lagging particle is just leaving one end of the path, say the left end, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?
 
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  • #3
ehild said:
"How far apart are they 0.46 s after the lagging particle leaves one end of the path?"
At t=0, the lagging particle is just leaving one end of the path, say the left end, where x1=0, and turning to move toward the other end at x2=1.5 m. What is the function x(t) then?
Then x(t)=1.5 this means cos(¢)=1
 
  • #4
i_hate_math said:
Then x(t)=1.5 this means cos(¢)=1
pls ignore my previous post,
x(0)=0=1.5cos(0+pi/2)
 
  • #5
i_hate_math said:
pls ignore my previous post,
x(0)=0=1.5cos(0+pi/2)
It is "simple harmonic motion along a common straight-line segment of length 1.5 m." That can mean from 0 to 1.5, for example, or from -0.75 to 0.75, as an other example. What is the amplitude?
 
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  • #6
ehild said:
It is "simple harmonic motion along a common straight-line segment of length 1.5 m." That can mean from 0 to 1.5, for example, or from -0.75 to 0.75, as an other example. What is the amplitude?
I see... 0.75 it is
 
  • #7
Yes. And what is the x(t) function, so as it is at the left end of the interval at t=0 and the velocity just turning to positive?
 
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  • #8
ehild said:
Yes. And what is the x(t) function, so as it is at the left end of the interval at t=0 and the velocity just turning to positive?
So for when its at the left end where x=0 at t=0,
x(0)=0.75cos(π/2) ===> ø=π/2
x(t)=0.75cos(wt+π/2)=0.75sin(wt)
and the other particle has
x'(t)=0.75cos(wt+π/5+π/2)=sin(wt+π/5)
 
  • #9
i_hate_math said:
So for when its at the left end where x=0 at t=0,
x(0)=0.75cos(π/2) ===> ø=π/2
x(t)=0.75cos(wt+π/2)=0.75sin(wt)
x(t) = 0.75sin(wt) varies between -0.75 and 0.75, and at t=0 it is at x=0. It is not one end of the line segment.
Decide where are the end of the light segment first.
 
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  • #10
ehild said:
x(t) = 0.75sin(wt) varies between -0.75 and 0.75, and at t=0 it is at x=0. It is not one end of the line segment.
Decide where are the end of the light segment first.
okay so at one end x=-0.75=0.75sin(wt), wt=3π/2
 
  • #11
Well, where are both particles 0.46 s later? How far are they?
 
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  • #12
ehild said:
Well, where are both particles 0.46 s later? How far are they?
X=sin(0.46w)=0.937282
 
  • #13
i_hate_math said:
X=sin(0.46w)=0.937282
The question was "How far apart are they 0.46 s after the lagging particle leaves one end of the path?". The lagging particle leaves the left end when wt = 3π/2. Where is it 0.46 s later?
 
  • #14
when it leaves the left end wt=3π/2 ==> t=3π/2w
0.46s later t'=3π/2w + 0.46
the position is x=sin(wt')=sin(3π/2+0.46w)
 
  • #15
i_hate_math said:
when it leaves the left end wt=3π/2 ==> t=3π/2w
0.46s later t'=3π/2w + 0.46
the position is x=sin(wt')=sin(3π/2+0.46w)
Yes, and what is the position of the other particle at the same time? What is the distance between them?
 
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  • #16
ehild said:
Yes, and what is the position of the other particle at the same time? What is the distance between them?
the other particle has position x=sin(3π/2+0.46w+π/5) since they differ in phase by π/5
the distance would be the difference between the two x values
 
  • #17
Yes, but you forgot the amplitude from both x-s.
Calculate the distance.
 
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  • #18
ehild said:
Yes, but you forgot the amplitude from both x-s.
Calculate the distance.
oh yes it should be
d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895
 
  • #19
i_hate_math said:
oh yes it should be
d=0.75sin(3π/2+0.46w+π/5)-0.75sin(3π/2+0.46w)=0.363261895
OK, it is correct.
Now question b) :)
 

What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around a central equilibrium point, with a restoring force acting towards the equilibrium point.

What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x = A sin(ωt + φ), where x is the displacement from the equilibrium point, A is the amplitude of the oscillation, ω is the angular frequency, and φ is the phase constant.

What is the difference between simple harmonic motion and periodic motion?

Simple harmonic motion is a type of periodic motion, meaning that it repeats itself over a specific time period. However, not all periodic motions are simple harmonic, as they may have different amplitudes or frequencies.

What factors affect the period of a simple harmonic motion?

The period of a simple harmonic motion is affected by the mass of the object, the spring constant, and the amplitude of the oscillation. As these factors change, the period of the motion will also change.

How is simple harmonic motion used in real life?

Simple harmonic motion is used in many real-life situations, such as in pendulum clocks, musical instruments, and even in the motion of planets around the sun. It is also used in engineering and physics to model and analyze various systems.

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