# Simple harmonic motion equations as a function of time

• zilex191
In summary: 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...f your system.In summary, when conducting a mass-spring experiment to determine the effects of stiffness and mass on frequency of oscillation, the displacement, velocity, and acceleration of the mass can be described by the equations x(t)=Acos(ωt+φ), v(t)=−Aωsin(ωt+φ), and a(t)=−Aω^2cos(ωt+φ), respectively. The values of A and φ can be determined based on the initial conditions, such as
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 ?

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.

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
kuruman said:
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

zilex191 said:
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.

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.

Last edited:
kuruman said:
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?

zilex191 said:
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
$$0 = A \cos{(\varphi)}$$
$$-0.5\text{ms}^{-1} = -A\omega \sin{(\varphi)}$$

zilex191 said:
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.

## What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium and is directed towards the equilibrium position.

## What is the equation for simple harmonic motion as a function of time?

The equation for simple harmonic motion as a function of time is x(t) = A*cos(ωt + φ), where x is the displacement from equilibrium, A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle.

## How do you calculate the period of simple harmonic motion?

The period of simple harmonic motion can be calculated using the equation T = 2π/ω, where T is the period and ω is the angular frequency.

## What is the relationship between frequency and period in simple harmonic motion?

The frequency of simple harmonic motion is equal to the reciprocal of the period, f = 1/T. This means that as the period increases, the frequency decreases, and vice versa.

## Can the simple harmonic motion equation be used for any type of periodic motion?

No, the simple harmonic motion equation can only be used for periodic motion where the restoring force is directly proportional to the displacement from equilibrium. If this condition is not met, a different equation must be used to describe the motion.

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