Simple harmonic motion equations as a function of time

[zilex191]

I conducted a mass-sprig experiment to see how stiffness of a spring and mass affect the frequency of oscillation. In addition to this to this i have to plot a graph to show displacement,velocity and acceleration of the mass as a function of time.From my research online

For the displacement as a function of time:
x(t)=x*cos(w*t)

For the velocity as a function of time(Deriving the above):
v(t)=x*w*sin(w*t)

For the acceleration as a function of time(Deriving the above):
a(t)=-x*w^2*cos(w*t)

But when i loot at other sources it shows different equations (such as instead of cos its sin).
For the displacement as a function of time:
x(t)=x*sin(w*t)

For the velocity as a function of time(Deriving the above):
v(t)=x*w*cos(w*t)

For the acceleration as a function of time(Deriving the above):
a(t)=-x*w^2*sin(w*t)

My question is what formula do i use ?

 

[kuruman]

The formula that you use depends on what you are trying to describe, namely what is the displacement of the mass at t = 0 and what is its velocity. These are the so-called initial conditions.

 

[Gaussian97]

You need to use, for position
$$x(t) = A \cos{(\omega t + \varphi)}$$
where $A>0$ is called the "amplitude" and tells you the maximum distance to the equilibrium, $\omega$ is the "angular frequency" and tells you how many oscillations you do in $2\pi$ seconds and $\varphi\in [0,2\pi)$ is called "initial phase" and essentially gives you the information on what is the initial position and initial velocity.
Differentiating you get:
$$v(t) = -A\omega \sin{(\omega t + \varphi)}, \qquad a(t) = -A\omega^2 \cos{(\omega t + \varphi)}$$

Note that your first set of equations is putting $\varphi=0$, and the second one is putting $\varphi=\frac{3\pi}{2}$.

 

[zilex191]

The formula that you use depends on what you are trying to describe, namely what is the displacement of the mass at t = 0 and what is its velocity. These are the so-called initial conditions.

Displacement at of the mass at t=0 is the maximum displacement which is 0.05 meters

 

[Gaussian97]

Displacement at of the mass at t=0 is the maximum displacement which is 0.05 meters

Then you must use (see my previous post)
$A=0.05 \text{m}$
$\phi = 0$
Although I would recommend you to try to figure out the values of $A$ and $\phi$ with your data because there are always some errors in setting the initial conditions.

 

[kuruman]

Then the expression to use is $x(t)=0.05~(\mathrm{m})\cos(\omega t)$. How do I know? Because at $t=0$ the expression gives $x(0)=0.05~(\mathrm{m})\cos(0)=0.05~\mathrm{m}.$

More generally, if the mass at $t=0$ is at $x(0)=x_0$ and has velocity $v(0)=v_0$, the position at any time $t$ is given by $x(t)=x_0\cos(\omega t)+\dfrac{v_0}{\omega}\sin(\omega t)$. Note that the expressions provided by @Gaussian97 in #3 are also correct but, in my opinion, less transparent in the general case.

 

[zilex191]

Then the expression to use is $x(t)=0.05~(\mathrm{m})\cos(\omega t)$. How do I know? Because at $t=0$ the expression gives $x(0)=0.05~(\mathrm{m})\cos(0)=0.05~\mathrm{m}.$

Thank you very much for your replies@kuruman @Gaussian97.
But in this case
Consider a body weighing 100 N suspended from a spring of constant k = 220 . At time t = 0, it has a downward velocity of 0.5 m.s-1 as it passes through the position of static equilibrium.

So i would use x(t)=Acos(ωt+φ) to work out the displacement x as a function of time, where x is measured from the position of static equilibrium?

 

[Gaussian97]

Thank you very much for your replies@kuruman @Gaussian97.
But in this case
Consider a body weighing 100 N suspended from a spring of constant k = 220 . At time t = 0, it has a downward velocity of 0.5 m.s-1 as it passes through the position of static equilibrium.

So i would use x(t)=Acos(ωt+φ) to work out the displacement x as a function of time, where x is measured from the position of static equilibrium?

Yes, with $k$ and $m$ you can compute $\omega$, then you need to solve the system of equations
$$0 = A \cos{(\varphi)}$$
$$-0.5\text{ms}^{-1} = -A\omega \sin{(\varphi)}$$

 

[kuruman]

Thank you very much for your replies@kuruman @Gaussian97.
But in this case
Consider a body weighing 100 N suspended from a spring of constant k = 220 . At time t = 0, it has a downward velocity of 0.5 m.s-1 as it passes through the position of static equilibrium.

So i would use x(t)=Acos(ωt+φ) to work out the displacement x as a function of time, where x is measured from the position of static equilibrium?

Note that I edited my previous post and gave you a general equation to describe your situation. It might be instructive to do it the way that @Gaussian97 suggests and then redo it the way I suggest.