Maximizing Displacements in SHM: Is ωt = -Φ/2 the Correct Value?

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

The displacements of the two particles are given as

y₁ = Asin(ωt)
y₂ = Asin(ωt+Φ)

y₂ - y₁ = Asin(ωt+Φ) -Asin(ωt)

y₂ - y₁ = 2Acos[(2ωt+Φ)/2]sin(Φ/2)

Maximum value of displacements between particles is given as 3A/2 . This will happen when cos(2ωt+Φ) = 1 or ωt = -Φ/2

When displacement is maximum , then 3A/2 = 2Asin(Φ/2)

Φ = sin^-1(3√7/8)

Is that correct till know ?

I think value of ωt = -Φ/2 is not correct as it gives opposite signs in displacements .

 

[phoenix95]

I think value of ωt = -Φ/2 is not correct as it gives opposite signs in displacements .

Having opposite signs makes sense, so when you calculate the difference it would be something like y₁-(-y₂) resulting a maximum displacement(The particles are further apart when they are on the either side of equilibrium position, compared to when they are on (one and)the same side of the equilibrium position)

That still leaves out the final part of the problem(calculating the displacement when it is equal). Do you need any help in how to proceed?

 

[Jahnavi]

How should I find the time at which the particles have the same displacement and hence find this displacement ?

 

[phoenix95]

y2 - y1 = 2Acos[(2ωt+Φ)/2]sin(Φ/2)

make use of this equation.

If the displacements are equal then what happens to the LHS? And so what can you deduce about the RHS?

 

[PeroK]

Homework Statement

View attachment 218454

Homework Equations

The Attempt at a Solution

The displacements of the two particles are given as

y₁ = Asin(ωt)
y₂ = Asin(ωt+Φ)

y₂ - y₁ = Asin(ωt+Φ) -Asin(ωt)

y₂ - y₁ = 2Acos[(2ωt+Φ)/2]sin(Φ/2)

Maximum value of displacements between particles is given as 3A/2 . This will happen when cos(2ωt+Φ) = 1 or ωt = -Φ/2

When displacement is maximum , then 3A/2 = 2Asin(Φ/2)

Φ = sin^-1(3√7/8)

Is that correct till know ?

I think value of ωt = -Φ/2 is not correct as it gives opposite signs in displacements .

Where did $\sqrt{7}$ come from?

You can continue with the same ideas to find when $y_2 = y_1$ now that you know $\phi$.

 

[Jahnavi]

make use of this equation.

If the displacements are equal then what happens to the LHS? And so what can you deduce about the RHS?

I get t = π/(2ω) - Φ/(2ω) , t is the time when displacements are equal .

Is that right ?

 

[phoenix95]

I get t = π/(2ω) - Φ/(2ω)

Is that right ?

You are on the right track.

Now since you got t, you can use these

y1 = Asin(ωt)
y2 = Asin(ωt+Φ)

to get the displacement.

I would recommend using the second equation(Try with both equations and find out why).

 

[PeroK]

I get t = π/(2ω) - Φ/(2ω) , t is the time when displacements are equal .

Is that right ?

Yes, although I would say $\omega t = \pi/2 - \phi/2$ or $\omega t = 3\pi/2 - \phi/2$

 

[Jahnavi]

I am getting value of displacement = Acos(Φ/2) , where Φ = sin^-1(3√7/8) .

Is that right ?

 

[Jahnavi]

@phoenix95 , but the answer given is A(√7/4)

 

[phoenix95]

but the answer given is A(√7/4)

I don't think that is correct

Because if you take the displacement(from the answer given), and substitute it in this:

y1 = Asin(ωt)

it would give you ωt=sin^-1(√7/4)

If you take the above value and the phase you calculated(which I think is correct), and substitute in the equation below:

y2 - y1 = 2Acos[(2ωt+Φ)/2]sin(Φ/2)

you won't get y₂-y₁=0.

 

[Jahnavi]

Where did $\sqrt{7}$ come from?

Φ = 2sin^-1(3/4) = sin^-1(3√7/8) .

@PeroK , do you think answer calculated in post#9 is correct ?

 

[PeroK]

@phoenix95 , but the answer given is A(√7/4)

This is correct. Your calculation of $\phi$ is wrong, as I tried to point out above.

 

[Jahnavi]

This is correct. Your calculation of $\phi$ is wrong, as I tried to point out above.

When displacement is maximum , then 3A/2 = 2Asin(Φ/2)

Is this correct ?

 

[PeroK]

When displacement is maximum , then 3A/2 = 2Asin(Φ/2)

Is this correct ?

Yes.

 

[Jahnavi]

Yes.

Then I am getting the same results as in post#12 . Please point out the error .

 

[PeroK]

I am getting value of displacement = Acos(Φ/2) , where Φ = sin^-1(3√7/8) .

Is that right ?

Okay, this is correct. The problem is how you get from this to your answer. I just used:

$\phi/2 = \sin^{-1}(3/4)$

I see what you did now. You used the double angle formula to get $\phi = \sin^{-1}(3 \sqrt{7}/8)$. I see now. That's right, but it just complicates things. Why not stick with the equation for $\phi/2$?

 

[PeroK]

This is correct. Your calculation of $\phi$ is wrong, as I tried to point out above.

Sorry, your calculation for $\phi$ was correct. I didn't do that calculation at all and when I checked it quickly, I got it wrong! Apologies for the confusion.

 

[Jahnavi]

Ok .

Now , I am getting the answer A(√7/4) . Is this what you are also getting ?

 

[PeroK]

Ok .

Now , I am getting the answer A(√7/4) . Is this what you are also getting ?

Yes.

 

[PeroK]

PS in fact, now that the problem is done, there was no need to use the inverse trig functions at all.

The first equation you get is that $\sin(\phi/2) = 3/4$. Just leave it like that!

And, the second equation you get is for the equal displacements:

$y_1 = y_2 = \pm A \cos(\phi/2)$

 

[Jahnavi]

@PeroK ,

As a cross check to verify whether the result A(√7/4) is correct , I put this value in place of y₁ . From this I get ωt = sin^-1(√7/4) .

Now when I put this value of ωt and value of Φ = sin^-1(3√7/8) , I should get y₂ = A(√7/4) .

But instead I am getting y₂ = A(5√7)/16 .

May be I have made some calculation mistake .

Could you please verify .

 

[PeroK]

@PeroK ,

As a cross check to verify whether the result A(√7/4) is correct , I put this value in place of y₁ . From this I get ωt = sin^-1(√7/4) .

Now when I put this value of ωt and value of Φ = sin^-1(3√7/8) , I should get y₂ = A(√7/4) .

But instead I am getting y₂ = 5√7/16 .

May be I have made some calculation mistake .

Could you please verify .

$\sin(\phi/2) = 3/4 \Rightarrow \phi/2 = 0.848$

$\omega t_0 = \pi/2 - \phi/2 = 0.723$

$y_1 = A\sin(\omega t_0) = 0.661A$

$y_2 = A\sin(\omega t_0 + \phi) = A \sin(2.419) = 0.661A$

 

[Jahnavi]

OK. Thanks !

But I am not supposed to use a calculator .So ,I need other ways to check my work . Please see this .

y₂ = Asin(ωt+Φ) ,

Put ωt = sin^-1(√7/4) and Φ = sin^-1(3√7/8) on the RHS .

Using the formula sin^-1x+sin^-1y = sin^-1[x√(1-y²)+y√(1-x²)]

I get y₂ = A(5√7)/16 .

Why am I not getting the exact result ?

 

[PeroK]

Let's do the general problem where the maximum displacement is $\alpha A$, where $0 \le \alpha \le 2$.

Now, we don't get tangled up with all those surds!

$\sin(\phi/2) = \alpha/2, \ \ \cos(\phi/2) = \sqrt{1 - \alpha^2 /4}$

$\omega t_0 = \pi/2 - \phi/2, \ 3\pi/2 - \phi/2$

$y_1 = A \sin(\omega t_0) = A\sin(\pi/2 - \phi/2), \ A\sin(3\pi/2 - \phi/2) = \pm A\cos(\phi/2)$

$y_2 = A \sin(\omega t_0 + \phi) = A\sin(\pi/2 + \phi/2), \ A\sin(3\pi/2 + \phi/2) = \pm A\cos(\phi/2) \ \$ (this is now a cross check, done algebraically, regardless of whether you know $\phi$ or not!)

In this case, $\alpha = 3/2$, hence $\cos(\phi/2) = \sqrt{7}/4$

And no calculator needed.

 

[ehild]

OK. Thanks !

But I am not supposed to use a calculator .So ,I need other ways to check my work . Please see this .

y₂ = Asin(ωt+Φ) ,

Put ωt = sin^-1(√7/4) and Φ = sin^-1(3√7/8) on the RHS .

Using the formula sin^-1x+sin^-1y = sin^-1[x√(1-y²)+y√(1-x²)]

I get y₂ = A(5√7)/16 .

Why am I not getting the exact result ?

You make it too complicated. You got that $\omega t = \pi/2 - \phi/2$. $y_2 = A\sin(\pi/2 +\phi/2) = A \cos (\phi/2)$.

 

[Jahnavi]

You make it too complicated. You got that $\omega t = \pi/2 - \phi/2$. $y_2 = A\sin(\omega t +\phi/2) = A \cos (\phi/2)$.

y₂ = Asin(ωt+Φ) not Asin(ωt+Φ/2)

 

[ehild]

y₂ = Asin(ωt+Φ) not Asin(ωt+Φ/2)

Thanks, it should have been π/2+Φ/2. I corrected it.

 

[Jahnavi]

I agree with both of you that I have made things complicated . Sorry !

Still , what is the mistake in my work in post#24 ?

 

[ehild]

OK. Thanks !

But I am not supposed to use a calculator .So ,I need other ways to check my work . Please see this .

y₂ = Asin(ωt+Φ) ,

Put ωt = sin^-1(√7/4) and Φ = sin^-1(3√7/8) on the RHS .

Using the formula sin^-1x+sin^-1y = sin^-1[x√(1-y²)+y√(1-x²)]

I get y₂ = A(5√7)/16 .

Why am I not getting the exact result ?

cos(Φ) should be -1/8.
sin (Φ/2) = 3/4 cos (Φ/2) =√7/4 --> Φ/2=48.6°, Φ=97.2 ° cos (Φ) is negative.

 

[Jahnavi]

Thank you very much @PeroK and @ehild .