# 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 ?

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Gold Member
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.

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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}##.

• vanhees71
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

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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.

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Gold Member
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.

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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?

Homework Helper
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
$$0 = A \cos{(\varphi)}$$
$$-0.5\text{ms}^{-1} = -A\omega \sin{(\varphi)}$$