Solve the Mistake: Simple Harmonic Motion

[zorro]

Homework Statement

A particle executes simple harmonic motion along a straight line with mean position at x=0, period 20s and amplitude 5cm. The (shortest) time in seconds taken by the particle to got from x=4cm to x=-3cm is?

The Attempt at a Solution

Let the equation be x=Asin(wt+Φ)

At t=0, let x=4 so that sinΦ=4/5
At t=t, putting x=-3 we get wt=sin^-1(-3/5) - sin^-1(4/5)

Solving for t we get t=-5s

Where is my mistake?

 

[SammyS]

Homework Statement

A particle executes simple harmonic motion along a straight line with mean position at x=0, period 20s and amplitude 5cm. The (shortest) time in seconds taken by the particle to go from x=4cm to x=-3cm is?

The Attempt at a Solution

Let the equation be x=Asin(wt+Φ)

At t=0, let x=4 so that sinΦ=4/5
At t=t, putting x=-3 we get wt=sin^-1(-3/5) - sin^-1(4/5)

Solving for t we get t=-5s

Where is my mistake?

Are you troubled by the time being negative, or by some other aspect of your solution?

If it takes -5s for the particle to go from x=4cm to x=-3cm, then it takes 5s for the particle to go from x=-3cm to x=4cm.

Starting at t=0, the particle goes from x=4 to x=5 to x=4 to x=0 to x=-3 to x=-5 to x=-3 to x=0 to x=4 to ... You want the time it takes to go from x=4 (the second time it's at x=4, highlighted) to x=-3 (also highlighted)

You have chosen Φ so that at t=0, the particle is at x = 4. It looks like you have implicitly taken Φ = sin^-1(4/5). This assumes that v is positive at t=0.

If Φ is chosen so that at t=0, the particle is at x = 4, but v is negative, then Φ = π - sin^-1(4/5).

Added in Edit:
(This is my (2¹⁰)th post on the PF site.)

 

[ehild]

Remember that sinΦ=a has two solutions between 0 and 2pi.

ehild

 

[zorro]

Are you troubled by the time being negative?

Yes.

You have chosen Φ so that at t=0, the particle is at x = 4. It looks like you have implicitly taken Φ = sin^-1(4/5). This assumes that v is positive at t=0.

How?
Are you trying to say that my equation should be x=Asin(wt-Φ)?

 

[ehild]

See attached picture. What is the shortest time to get from x=4 to x=-3?

ehild

 

[zorro]

0.012s

 

[ehild]

Is this the solution of your problem?

ehild

 

[zorro]

No it is not.

 

[rl.bhat]

The equation is x = Asin(ωt + φ)
Two equations for to positions are
4 = 5*sin(ωt1 + φ), the angle lies between π/2 and π.
-3 = 5*sin(ωt2 + φ), the angle lies between π and 3π/2.
Now can you find t1 - t2?

 

[zorro]

I don't want to get into the trouble of finding t1-t2. I want to find out 't' directly by choosing a proper equation.

 

[gneill]

Have a look at the attached diagram showing the relevant angles on a 'unit circle' scaled to 5cm. The linear motion of the problem is the projection onto the y-axis of a radius vector going counterclockwise around the circle with a 20 second period.

You're looking for the shortest time between +4 and -3, which corresponds to the points t1 and t2 on the diagram. The problem is, when you use the arcsin function it returns 'principal angles' that lie in the first or fourth quadrant. That is, it gives θ1 and θ2 rather than φ1 and φ2.

No problem though! By symmetry, the time from points t1 to t2 is the same as the time from points a to b. So your desired formula becomes:

ΔT = (arcsin(4/5) - arcsin(-3/5))*(20s/2π)

 

[ehild]

I don't want to get into the trouble of finding t1-t2. I want to find out 't' directly by choosing a proper equation.

Then find a proper phase constant, an angle which sine is 0.8. But there are two such angles between 0 and 2pi, as you can see in my picture, as sin(φ)=sin(π-φ). You can place t=0 at both of them, and also you can find two angles which sine is -0.3. So you have to calculate the time with both phase constants and choose the shortest.
My way is to choose zero phase constant and find the phases both at x=0.8 and x=-0.3. As time goes forward, all of them have to be positive.

ehild

 

[ehild]

I misread the problem in my former post, and took the frequency equal to 20, instead of the time period. I replaced the picture with the correct one (I hope!)

ehild

 

[zorro]

Reading the last 3 posts, I understood something. This is my analysis of the question-

1. Using sine function (which is a bit confusing)

x=Asin(wt+φ)
at t=0, sinφ=4/5
But the particle goes from 4 to 5 then comes back. So we have to consider another solution of the equation
i.e. sin(π-φ)=4/5 so that the particle travels from x=4 to x=-3
φ=π-sin^-1(4/5)

-3=5sin(wt+φ)
sin(π+wt+φ)=3/5
wt=sin^-1(3/5)-π-φ

on solving t=+5s

2. Using cosine function (easier way)

x=5cos(wt+φ)
φ=cos^-1(4/5)

-3=5cos(wt+φ)
wt=cos^-1(-3/5)-φ=π-cos^-1(3/5)-cos^-1(4/5)

Again t=+5s.

I don't understand one thing, why there is no need to consider other solutions with cosine function?

 

[SammyS]

Yes, but using the using the sin function helped you to understand some of the possible difficulties you can run into with this type of problem.

I was waiting for you to get your answer with the sin function, before mentioning that the cos function would give the answer immediately -- but not the understanding.

Good Job !

 

[rl.bhat]

I don't understand one thing, why there is no need to consider other solutions with cosine function?
Because the cosine function starts from the extreme position of the oscillation.So in a single swing particle goes from +4 cm to - 3 cm. Hence the solution is simple.
Phase is the initial position of the oscillating particle when you start your observation. In the problem, it is not given whether the particle is going towards or away from the equilibrium position. But the time taken by the particle is minimum when the particles moves directly from 4 cm to -3 cm.