Simple Harmonic Motion question (Need clarification)

[Augustine Duran]

Homework Statement

The point of the needle of a sewing machine moves in SHM along the x-axis with a frequency of 2.5 Hz. At t=0 its position and velocity components are +1.1 cm and -15 cm/s, respectively.
(a) Find the acceleration component of the needle at t=0
(b) write an equation giving the position at the point as a function of time.

I just want to state that I already know the answers, my question is concerning the value of the Amplitude in both (a) and (b).

My first question:
So for part (a) the answer for acceleration is -2.71 m/s². This is found by using the equation

a(t) = -ω²A = -2.71 m/s²

where ω = 5π and A = 0.011 m. Why is the amplitude 0.011m? In the question it states that at t=0 its at a certain position , 1.1cm, with a negative velocity, if it has a velocity then its not at rest, and if its not at rest that means its not at its max position away from the equilibrium point, which from what I understand is defined as the Amplitude. So why is 1.1 cm used as the amplitude in this case?

My 2nd question

In part (b) the answer for the position equation is:

x(t) = (0.0146)cos[(15.7 rad/s)t + 0.715 rad]

So in this case A = 0.0146. This is found by using the equation

A = [x_o² + v_o²/ω²]^½

So why does this have a different Amplitude from in part (a). Should the value for A be the same in both cases?

 

[kuruman]

You got the wrong equation to consider. The correct equation is a(t) = -ω²x(t). If the acceleration were -ω²A, it would be constant, would it not? You are confusing instantaneous position x(t) with amplitude A. One depends on time, the other does not.

 

[ehild]

Homework Statement

The point of the needle of a sewing machine moves in SHM along the x-axis with a frequency of 2.5 Hz. At t=0 its position and velocity components are +1.1 cm and -15 cm/s, respectively.
(a) Find the acceleration component of the needle at t=0
(b) write an equation giving the position at the point as a function of time.

I just want to state that I already know the answers, my question is concerning the value of the Amplitude in both (a) and (b).

My first question:
So for part (a) the answer for acceleration is -2.71 m/s². This is found by using the equation

a(t) = -ω²A = -2.71 m/s²

where ω = 5π and A = 0.011 m. Why is the amplitude 0.011m? In the question it states that at t=0 its at a certain position , 1.1cm, with a negative velocity, if it has a velocity then its not at rest, and if its not at rest that means its not at its max position away from the equilibrium point, which from what I understand is defined as the Amplitude. So why is 1.1 cm used as the amplitude in this case?

My 2nd question

In part (b) the answer for the position equation is:

x(t) = (0.0146)cos[(15.7 rad/s)t + 0.715 rad]

So in this case A = 0.0146. This is found by using the equation

A = [x_o² + v_o²/ω²]^½

So why does this have a different Amplitude from in part (a). Should the value for A be the same in both cases?

Yes, the amplitude A should be the same for both questions (a) and (b). How do you know that the value of the acceleration was calculated using A=1.1 cm? By the way, a(0)=-2.71 m/s² is correct.

 

[Augustine Duran]

You got the wrong equation to consider. The correct equation is a(t) = -ω²x(t). If the acceleration were -ω²A, it would be constant, would it not? You are confusing instantaneous position x(t) with amplitude A. One depends on time, the other does not.

Okay looking through my book again I see that they subtly mention the equation a(t) = -ω²x(t). Okay so x(t) in that equation would equal to:

x(t) = Acos(ωt), correct?

 

[kuruman]

x(t) = Acos(ωt), correct?

Incorrect. The expression you quoted predicts that at t = 0 the needle is at maximum displacement. You need to find an equation that puts the needle at 1.1 cm with velocity -15 cm/s when t = 0.

 

[Augustine Duran]

sry I am at a loss here. the only equation i can find that relates velocity and position is V_o/x_o = -ωtanΦ

 

[ehild]

The general form of SHM displacement is x(t)=Acos(ωt+Φ). What is the velocity then?

 

[kuruman]

sry I am at a loss here. the only equation i can find that relates velocity and position is Vo/xo = -ωtanΦ

What a meant was find an equation for the position and use it to find an equation for the velocity. SOrry if I confused you.

x(t) = (0.0146)cos[(15.7 rad/s)t + 0.715 rad]

You found the above equation but it does not quite do it. Plug in and you will see. You need to fix it. Knowing the position as a function of time allows you to find everything else as a function of time. However, for part (a) you can use a(0) = -ω² x(0) directly.

 

[Augustine Duran]

The general form of SHM displacement is x(t)=Acos(ωt+Φ). What is the velocity then?

v(t) = -ωAcos(ωt+Φ)

What a meant was find an equation for the position and use it to find an equation for the velocity. SOrry if I confused you.

x(t) = (0.0146)cos[(15.7 rad/s)t + 0.715 rad]

You found the above equation but it does not quite do it. Plug in and you will see. You need to fix it. Knowing the position as a function of time allows you to find everything else as a function of time. However, for part (a) you can use a(0) = -ω² x(0) directly.

i plugged it in and got 0.011m

 

[kuruman]

Sorry, I was incorrect when I said the equation does not quite do it. It is correct. So, can you answer questions (a) and (b) now? Is there something you still don't understand?

 

[Augustine Duran]

well here's what i initially tried to do. (hopefully I am not being too redundant here)

When it asked me to find the acceleration component i immediately thought about using,
a(t) = -ω²Acos(ωt+Φ)
from there i would set Φ=0 and use the equation A = (x_o²+v_o²/ω²)^½ to find A giving me a(t) = -ω²A . I would just then plug in my values for A and ω, but i got the answer wrong

i guess if you could explain one more time the difference between the equation i used here, and the equation you gave me which was a(t) = -ω²x

 

[kuruman]

i guess if you could explain one more time the difference between the equation i used here, and the equation you gave me which was a(t) = -ω²x

Start with the general form: $x(t)=A\cos(\omega t+\varphi)$
1. Take the derivative to find the velocity: $v(t) = -\omega A sin((\omega t+\varphi)$
2. Take the derivative once more to find the acceleration: $a(t) = -\omega^2 A cos((\omega t+\varphi)$
3. Note that from the general form this can be written as $a(t) = -\omega^2 x(t)$

The hallmark of SHM is that the acceleration is always proportional to the displacement, the constant of proportionality being the negative of the frequency squared.

 

[Augustine Duran]

ok i think i see it now

x was already defined at 1.1cm at t=0 so , Acos(ωt+Φ) = 1.1cm and substitute into a(t) = -ω²Acos(ωt+Φ)

That also makes sense that its proportional to the displacement considering there's no force at the equilibrium point x=0 then a=0. units check out also.

thanks for your patience i appreciate it

 

[kuruman]

x was already defined at 1.1cm at t=0 so , Acos(ωt+Φ) = 1.1cm and substitute into a(t) = -ω2Acos(ωt+Φ)

Just for completeness, I should add that defining x(0) = 1.1 cm is not enough because the oscillator could be moving either away from the origin (positive velocity) or towards the origin (negative velocity.) You need the additional information that v(0) = -15 cm/s to specify the needed parameters. This allows setting up a system of two equations and two unknowns, the unknowns being A and Φ.

 

[ehild]

ok i think i see it now

x was already defined at 1.1cm at t=0 so , Acos(ωt+Φ) = 1.1cm and substitute into a(t) = -ω²Acos(ωt+Φ)

The equation in red is wrong. The left-hand side depends on time, it is not equal to a single value.
Correctly, x(0) = 1.1 cm and v(0)=-15 cm/s means Acos(Φ) = 1.1cm and -Aω sin(Φ) = -15 cm/s. Dividing the two equations you get tan(Φ)=v₀/(x(0)ω).

 

[naMuRaS]

well here's what i initially tried to do. (hopefully I am not being too redundant here)

When it asked me to find the acceleration component i immediately thought about using,
a(t) = -ω²Acos(ωt+Φ)
from there i would set Φ=0 and use the equation A = (x_o²+v_o²/ω²)^½ to find A giving me a(t) = -ω²A . I would just then plug in my values for A and ω, but i got the answer wrong

i guess if you could explain one more time the difference between the equation i used here, and the equation you gave me which was a(t) = -ω²x

Hi where do you get the equation for amplitude from? Cheers.

 

[kuruman]

Hi where do you get the equation for amplitude from? Cheers.

Please note that this thread is more than 5 years old and that @Augustine Duran has not been seen here since Mar. 30 2019. It is unlikely that you will get a response to your query, I am sorry to say. I cannot speak for Augustine, but I would hazard a guess and say that he got the amplitude by starting from the harmonic solution $x(t)=A\cos(\omega t+\phi)$. Then he took the time derivative to get $v(t)=-\omega A\sin(\omega t+\phi).$ Then he said, $$\begin{align} & x(0)=x_0=A\cos\phi \implies A\cos\phi=x_0 \nonumber \\ & v(0)=v_0=-\omega A \sin\phi \implies A\sin\phi=-\frac{v_0}{\omega} \nonumber\end{align}$$ Then, being a clever chap, he thought "why don't I square both sides and add the equations?"$$\begin{align} & A^2\cos^2\phi +A^2\sin^2\phi=x_0^2+\left(-\frac{v_0}{\omega}\right)^2 \nonumber \\
& A^2=x_0^2+\frac{v_0^2}{\omega^2}\implies A=\left(x_0^2+\frac{v_0^2}{\omega^2}\right)^{1/2}
\nonumber\end{align}$$ Cheers to you, too.

 

[naMuRaS]

Please note that this thread is more than 5 years old and that @Augustine Duran has not been seen here since Mar. 30 2019. It is unlikely that you will get a response to your query, I am sorry to say. I cannot speak for Augustine, but I would hazard a guess and say that he got the amplitude by starting from the harmonic solution $x(t)=A\cos(\omega t+\phi)$. Then he took the time derivative to get $v(t)=-\omega A\sin(\omega t+\phi).$ Then he said, $$\begin{align} & x(0)=x_0=A\cos\phi \implies A\cos\phi=x_0 \nonumber \\ & v(0)=v_0=-\omega A \sin\phi \implies A\sin\phi=-\frac{v_0}{\omega} \nonumber\end{align}$$ Then, being a clever chap, he thought "why don't I square both sides and add the equations?"$$\begin{align} & A^2\cos^2\phi +A^2\sin^2\phi=x_0^2+\left(-\frac{v_0}{\omega}\right)^2 \nonumber \\
& A^2=x_0^2+\frac{v_0^2}{\omega^2}\implies A=\left(x_0^2+\frac{v_0^2}{\omega^2}\right)^{1/2}
\nonumber\end{align}$$ Cheers to you, too.

Genius. Thanksss.

 

[malawi_glenn]

Note that this is a standard technique, whenever you have $\cos \phi = \alpha$ and $\sin \phi = \beta$ one should immediatly think "let's use $\cos^2 \phi + \sin^2 \phi = 1$" and thus get $\alpha^2 + \beta^2 = 1$