B Correct SHM Equation: Does € Matter?

[Curiosity_0]

A textbook I am using gives the basic eqn of motion of shm as follows :
X = Asin(wt + €)
V =Awcos(wt+€)
But other textbooks and online sources are interchanging sin and cos in above equations, so which is the correct one? Or does it depend on the phase constant €?

 

[topsquark]

A textbook I am using gives the basic eqn of motion of shm as follows :
X = Asin(wt + €)
V =Awcos(wt+€)
But other textbooks and online sources are interchanging sin and cos in above equations, so which is the correct one? Or does it depend on the phase constant €?

To paraphrase Gertrude Stein: A sine function is a sine function is a sine function. The best way to write the SHO solution is to let $X(t) = A ~sin( \omega t + \phi )$ where the phase angle $\phi$ is left to the boundary conditions. Yes, the phase angle makes the solution general to all SHO. Applications like a spring or pendulum tend to use cosine because we usually start the motion (t = 0) at an extreme extension and cos(0) = 1 is the maximum of the cosine function.

-Dan

 

[Dale]

which is the correct one

There is no correct or incorrect one. It is simply a matter of convention.

 

[malawi_glenn]

Sin and cos are related by a pi/2 phase. Both works

 

[PeroK]

A textbook I am using gives the basic eqn of motion of shm as follows :
X = Asin(wt + €)
V =Awcos(wt+€)
But other textbooks and online sources are interchanging sin and cos in above equations, so which is the correct one? Or does it depend on the phase constant €?

Note that $$\cos(wt + \phi) = \sin(wt + \phi + \frac \pi 2)$$In other words, every sine function is also a cosine function with a different phase and vice versa.

 

[kuruman]

Also note that, using a standard trig identity, $$\begin{align} & \cancel{A\cos(\omega t+\delta)=A\cos\delta \sin\omega t+A\sin\delta \cos\omega t} \nonumber \\ & A\sin(\omega t+\delta)=A\cos\delta \sin\omega t+A\sin\delta \cos\omega t \nonumber \end{align}$$With the definitions $$C_1\equiv A\cos\delta~;~~C_2\equiv A\sin\delta$$you have $$A\cos(\omega t+\delta)=C_1\sin\omega t + C_2\cos\omega t$$so the expressions are equivalent. Note that the amplitude and constant phase can be found from the definitions, $$A=\sqrt{C_1^2+C_2^2}~;~~\delta = \arctan\left(\frac{C_2}{C_1}\right)$$so you can go back and forth from one convention to the other.

On edit: Fixed wrong trig function on the LHS of the identity. Also, the phase has a sign ambiguity as noted in posts #7, #9 and #10.

 

[vanhees71]

It's more safe to use
$$\delta=\arccos \left (\frac{C_1}{\sqrt{C_1^2+C_2^2}} \right ) \mathrm{sign} \, C_2.$$

 

[Herman Trivilino]

Or does it depend on the phase constant €?

It does indeed. You can turn sine into cosine or vice-versa by changing the value of the phase constant.

Physically, this is equivalent to starting the oscillator in different positions and different velocities. In other words, when $t=0$ you can make $x$ have any value (in the closed interval $[-A, A]$) by adjusting the value of the phase constant.

 

[PeroK]

$$A\cos(\omega t+\delta)=C_1\sin\omega t + C_2\cos\omega t$$

I use:
$$A\sin x + B\cos x = sgn(A)\sqrt{A^2+B^2} \sin(x + \delta) \ \ \ (\delta = \tan^{-1}\big (\frac B A \big))$$And
$$A\sin x + B\cos x = sgn(B)\sqrt{A^2+B^2} \cos(x - \delta) \ \ \ (\delta = \tan^{-1}\big (\frac A B\big))$$

 

[vanhees71]

Also here, be careful with the use of the arctan-formula. You have to make sure to get the phase right. It's much easier to use the formula, adapted to your other conventions, given in #7. For this reason, when programming for that purpose you rather use the function atan2, rather than atan!

 

[PeroK]

Also here, be careful with the use of the arctan-formula. You have to make sure to get the phase right.

There's a proof of the latter identity here:

https://www.physicsforums.com/threads/trigonometric-identity-involving-sin-cos.1046567/#post-6812748

 

[vanhees71]

Sigh. I just want to say that you need to be careful with the arctan formula. The result is always a value in the interval $(-\pi/2,\pi/2)$. What you obviously need to get a unique map between the Cartesian components of an $\mathbb{R}^2$ vector and its polar coordinates is a value within an interval of the length $2 \pi$. Choosing the interval $(-\pi,\pi]$, you get $(x,y)=r(\cos \varphi,\sin \varphi)$, using
$$r=\sqrt{x^2+y^2}, \quad \varphi=\begin{cases} \arccos \left (\frac{x}{r} \right) \text{sign} \, y, &\text{for} \quad y \neq 0, \\ 0 & \text{for} \quad x>0, \quad y=0 \\ \pi & \text{for} \quad x<0, \quad y=0. \end{cases}$$