# Oscillatory motion (vertical spring-mass system)

• gramentz
In summary, Resonance in an un-damped spring-mass system will cause a variation in velocity and y-position according to the equation v= 5cos2t- 16y'.
gramentz
"Resonance in an un-damped spring-mass system"

If I have a force that is pushing upwards on a spring-mass system, and I basically have to find an equation that will give me the velocity and y-position for any given t, how much does that differ from the general form of Asin ($$\omega$$t +$$\phi$$)?

This is what I know:
m= 2
k= 32

Vertical position y of mass: dy/dt = v

Velocity of the mass: dv/dt = f(t)/M - (k/M) * y

I know the external force is f(t) is 10sin($$\omega$$t)

I know that y(0) and v(0) are zero.

I have to find the position and velocities when $$\omega$$= 2, 3, 3.5, 4, 4.5, 5 rad sec and I have to use 40,000 time steps (totaling 25 seconds) where each step is 0.000625.

I know that the "preferred" frequency of the system will be at 4 rad/ sec, using $$\omega$$= $$\sqrt{\frac{k}{m}}$$

This is where I am having a bit of trouble: I found that when I'm examining 2 rad/sec, that v'' = 5cos2t - 16y', or that v'' = 5cos2t-16v. What I'm trying to do, is find the velocity and position of this system for any given value of t.

I was told that I can use the "general" solution of this type: v= A cos 2t + B cos 4t + C sin 2t + D sin 4t.

Am I in the right direction? Any insight would be appreciated. Thanks in advance!

Is this being driven by an oscillitary force, such as F(t) = F0sin(ωt) ?
Is ω0=sqrt(k/m), where ω0 ≠ ω ?

Is this homework??

Bob S

No, this isn't homework.

Yes, the external force that is applied is given by 10sin(wt) where w will vary.

I am first examining what happens when w = 2 rad/sec.

I've been working on this and came up with the following at 2 rad/ sec (don't know if it is correct)

dv/ dt = 5sin(2t) - 16y

since y' = v, y= v^2/2

dv/ dt = 5sin(2t)- 8v^2

dv/dt +8v^2 = 5sin (2t)
8v^2 dv = 5sin (2t)dt

(8/3)v^3 = (-5/2)cos(2t)
v = cubed root of (-15/16 cos (2t))

Isn't sqrt(k/m) fixed at sqrt(32/2) = w0 = 4 radians per second?

y'' +w02y = (10/m) sin(wt) [STRIKE]- g[/STRIKE] where the right hand term is the driving term?

where w0= sqrt(k/m)

Note that the units on the left hand and right hand sides have to match.

Bob S

Last edited:
Hi gramentz-

Look at the thumbnail and see if it fits all conditions, including driving force and two initial conditions.

[added] Here is a thumbnail plot of the solution using w=3.5 radians per second, and w0 = 4 radians per second. Both the amplitude and derivative are zero at t=0.

Bob S

#### Attachments

• Forced_oscillator.jpg
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• Forced_oscillator_plot.jpg
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Last edited:
When we assume the general solution of the form y= csin(w0t) + dcos(w0t)...is this because all of the derivatives will take such form? Since dy/dt = v in the original equation I was given, that implies that step 7 in the thumbnail could also be v(t)? Thank you for helping me with this.

Hi gramentz-

The particular solution provides the long-term (steady state) solution to match the driving force, especially when there is damping. But the particular solution does not match the initial conditions; specifically y(0) = y'(0) = 0. To do this, we need a solution to [4] with two arbitrary constants that can match the particular solution [1-3] to the two initial conditions. So that is why a general solution of the form [5] was chosen. The final solution [8] matches the two initial conditions at t=0, AND the particular equation [1] at all times.

Bob S

Last edited:
It took me quite some time, but I finally understand what you wrote and how you arrived at your answers. Thanks for taking the time to help me out, I appreciate it.

## 1. What is oscillatory motion?

Oscillatory motion refers to the back and forth movement of an object around a central point or equilibrium position. This type of motion is characterized by the repetition of a specific pattern or cycle.

## 2. What is a vertical spring-mass system?

A vertical spring-mass system is a physical system in which a mass is attached to a spring and allowed to move vertically. The mass-spring system can exhibit oscillatory motion when the mass is displaced from its equilibrium position.

## 3. What factors affect the frequency of oscillations in a vertical spring-mass system?

The frequency of oscillations in a vertical spring-mass system is affected by the mass of the object, the stiffness of the spring, and the gravitational acceleration. The heavier the mass, the stiffer the spring, and the higher the gravitational acceleration, the higher the frequency of oscillations will be.

## 4. How is the period of oscillations related to the frequency of oscillations in a vertical spring-mass system?

The period of oscillations, which is the time it takes for one complete cycle of motion, is inversely related to the frequency of oscillations. This means that the higher the frequency, the shorter the period, and vice versa.

## 5. What is the equation for calculating the frequency of oscillations in a vertical spring-mass system?

The equation for calculating the frequency of oscillations in a vertical spring-mass system is f = 1 / (2π) * √(k/m), where f is the frequency, k is the spring constant, and m is the mass of the object attached to the spring.

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