Question regarding Simple Harmonic Motion

In summary, the problem asks to find the resultant path of a particle undergoing two perpendicular linear simple harmonic motions with equal amplitudes and angular frequencies ω and 2ω along the axes of X and Y respectively, given an initial phase difference of π/2 between them. This can be solved by using parametric equations and eliminating the ωt term, resulting in a Lissajous curve or a Lemniscate of Gerono, depending on the initial phases of the two motions.
  • #1
sankalpmittal
785
15

Homework Statement



The two linear simple harmonic motions of equal amplitudes , and angular frequencies ω and 2ω are imposed on a particle along the axes of X and Y respectively. If the initial phase difference between them is π/2 , then find the resultant path followed by the particle.


Homework Equations



http://en.wikipedia.org/wiki/Simple_harmonic_motion

The Attempt at a Solution



I tried solving question using x= a sinωt
And other is y=a sin(2ωt + π/2)
Solving them , I get the wrong answer in terms of x and y...

Please help !

Thanks in advance...
:smile:
 
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  • #2
Can you show how you went about "solving them"? We need to see your attempt if we're to know how to help.
 
  • #3
Hi sankalpmittal,

Show please what you have tried?
x=sin(ωt) and y=sin(2ωt+π/2) =cos(2ωt) is the parametric representation of a Lissajous curve. See http://en.wikipedia.org/wiki/Lissajous_curve

Try to eliminate the ωt term.

ehild
 
  • #4
ehild said:
Hi sankalpmittal,

Show please what you have tried?
x=sin(ωt) and y=sin(2ωt+π/2) =cos(2ωt) is the parametric representation of a Lissajous curve. See http://en.wikipedia.org/wiki/Lissajous_curve

Try to eliminate the ωt term.

ehild

Firstly I solved this question using :
a : Amplitude
x=a cosωt
y=a sin(2ωt+π/2)
y=a cos (2ωt)
y= a( cos2ωt-sin2ωt)
y= a(2cos2ωt-1)
y= a{(2x2/a2)-1)

I noticed that this does not match the answer given in my textbook...

Then I made a second attempt :
x=a sinωt
y=a sin (2ωt+π/2)
y=a cos(2ωt)
y= a(1-2sin2ωt)
y= a{1-(2x2/a2)}

Again this does also not match with the answer in my textbook...

Then I made the third attempt :

x= a cosωt
y= a cos(2ωt+π/2)
y= -asin(2ωt)
y= -2a sin(ωt)cos(ωt)
y=-2a √(1-sin2ωt) (x/a)
y=-2a √{1-(x2/a2)} (x/a)
On squaring both sides and simplifying , I got :

y= 4x2{1-(x2/a2)}

This answer matched with the answer given in my textbook...

I just wanted to know , where I went wrong in my first and second attempt...
 
  • #5
sankalpmittal said:
Firstly I solved this question using :
a : Amplitude
x=a cosωt
y=a sin(2ωt+π/2)
y=a cos (2ωt)

They are ins phase, instead of being shifted by pi/2.

sankalpmittal said:
Then I made a second attempt :
x=a sinωt
y=a sin (2ωt+π/2)
y=a cos(2ωt)
y= a(1-2sin2ωt)
y= a{1-(2x2/a2)}

Again this does also not match with the answer in my textbook...
.

That is correct, if x and y are as you assumed.

sankalpmittal said:
Then I made the third attempt :

x= a cosωt
y= a cos(2ωt+π/2)
y= -asin(2ωt)
y= -2a sin(ωt)cos(ωt)
y=-2a √(1-sin2ωt) (x/a)
y=-2a √{1-(x2/a2)} (x/a)
On squaring both sides and simplifying , I got :

y= 4x2{1-(x2/a2)}

This answer matched with the answer given in my textbook...

But this answer is wrong as y is not squared. It was correct before squaring.

If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases.


ehild
 
  • #6
ehild said:
They are ins phase, instead of being shifted by pi/2.

.

That is correct, if x and y are as you assumed.



But this answer is wrong as y is not squared. It was correct before squaring.

If you add two perpendicular vibrations you can have different y(x) curves, depending on the initial phases.


ehild

OK , so there was a typo. The answer I got in my third attempt was y2=4x2{1-(x2/a2)} , which matched with the answer given in my textbook. But unfortunately the answer which I got in my first and second attempt , did not match with that given in my textbook.

According to you , both the equations , in my second and third attempt are correct , yet they are different and only latter match with the answer in my textbook. How ?

Also what do you mean by "in phase and not shifted by pi/2" in my first attempt ? Can you explain a bit comprehensively ?
 
  • #7
Your first attempt was
x=a cosωt
y=a sin(2ωt+π/2) which is equivalent to
y=a cos (2ωt)I do not see any initial phase difference between x and y.
Your second attempt is true and so is the third one. Phase difference between x and y is not a clear concept. You get x,y curves of different form if x=sin(ωt), y=cos(2ωt) (y=1-2x2) and when x=cos(ωt) and y=sin(2ωt) (y2=4x2(1-x2))

ehild
 

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  • #8
sankalpmittal said:

Homework Statement



The two linear simple harmonic motions of equal amplitudes , and angular frequencies ω and 2ω are imposed on a particle along the axes of X and Y respectively. If the initial phase difference between them is π/2 , then find the resultant path followed by the particle.

Homework Equations



http://en.wikipedia.org/wiki/Simple_harmonic_motion
...
A better link to consider might be: http://en.wikipedia.org/wiki/Lisajous .

Also see: http://en.wikipedia.org/wiki/Lemniscate_of_Gerono .
 

What is Simple Harmonic Motion (SHM)?

Simple Harmonic Motion is a type of periodic motion in which an object moves back and forth in a straight line due to a restoring force, such as the force of gravity or a spring. It follows a specific pattern of acceleration and displacement and can be described by mathematical equations.

What is the equation for SHM?

The equation for SHM is x = A sin(ωt + φ), where x is the displacement of the object, A is the amplitude, ω is the angular frequency, and φ is the phase constant. This equation can also be written as x = A cos(ωt + φ) depending on the starting position of the object.

What is the relationship between period and frequency in SHM?

The period of SHM is the time it takes for one complete cycle of motion, while the frequency is the number of cycles per second. The relationship between period (T) and frequency (f) is given by the equation f = 1/T or T = 1/f. In other words, the period and frequency are inversely proportional.

What is the role of the restoring force in SHM?

The restoring force is responsible for bringing the object back to its equilibrium position in SHM. It acts in the opposite direction of the displacement, causing the object to oscillate back and forth around the equilibrium point. The magnitude of the restoring force increases as the displacement increases, resulting in a greater acceleration and velocity.

What are some real-life examples of SHM?

Some common examples of SHM include the motion of a pendulum, the oscillation of a mass attached to a spring, and the vibrations of a guitar string. Other examples include the motion of a swing, the vibrations of a tuning fork, and the movement of a diving board. SHM can also be observed in the motion of molecules and atoms in a solid or liquid.

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