How Do You Model the Motion of a Mass on a Spring?

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Homework Help Overview

The discussion revolves around modeling the motion of a mass attached to a vertical spring, focusing on the mathematical expressions for displacement, velocity, and acceleration as functions of time. Participants are analyzing the effects of initial conditions and the choice of trigonometric functions in the context of harmonic motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate trigonometric function to use for modeling the displacement, with some suggesting sine and others proposing cosine based on the initial conditions. There are calculations of angular frequency (ω) and attempts to derive expressions for displacement, velocity, and acceleration.

Discussion Status

Several participants have provided calculations and expressions, while others are questioning the assumptions made regarding the initial conditions and phase angle. There is a mix of approaches being explored, with some guidance offered on using a more general form for the displacement equation.

Contextual Notes

Participants note potential confusion regarding the wording of the problem, particularly about the timing of maximum displacement and the phase angle. There is an emphasis on inserting known values into the equations to derive further insights.

Kev1n
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1. A mass attached to the lower end of a vertical spring causes the spring to extend by 25 mm to its equilibrium position. The mass is then displaced a further 20 mm and released. A trace of the vibration and time measurements are taken. From these measurements it can be seen that the displacement from the equilibrium position is 19.2 mm when the time is 0.05 s.)
(a) Write the expression for the displacement of the mass as a function of time.
(b) Write the expression for the velocity of the mass as a function of time.
(c) Write the expression for the acceleration of the mass as a function of time.




2. a. Y= ASinWT, b. v= AwCoswT, c. a= -Aw^2Sin wT



3. I have calculated x = 0.0192, t = 0.05, w = 19.8, A = 0.02

Attempt

a. Y = 20Sin19.8t
b. v = 396Cos19.8t
c. a= -7840.8Sin19.8t

Any advice on where I am or what I am doing wrong appreciated, thanks
 
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Since the mass is released at its maximum displacement from equilibrium, you'd be better off using cosine instead of sine to describe the displacement.

How did you compute ω?
 
Doc Al said:
Since the mass is released at its maximum displacement from equilibrium, you'd be better off using cosine instead of sine to describe the displacement.

How did you compute ω?

W = sqrt K/M = sqrt 9810/25 = W^2 = 9810/25 = 19.8

I am actually quite lost as to what I have to achieve
 
Kev1n said:
W = sqrt K/M = sqrt 9810/25 = W^2 = 9810/25 = 19.8
Good.
I am actually quite lost as to what I have to achieve
I presume the position and time were given so you can calculate the phase at t = 0. (I guess it isn't at maximum displacement at t = 0, as I had first thought. A bit strangely worded, I'd say.)

Use a more general form for the displacement; something like y = A sin(ωt + Φ).
 
Doc Al said:
Good.

I presume the position and time were given so you can calculate the phase at t = 0. (I guess it isn't at maximum displacement at t = 0, as I had first thought. A bit strangely worded, I'd say.)

Use a more general form for the displacement; something like y = A sin(ωt + Φ).

Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0
 
Kev1n said:
Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0

Then work out a figure for each epression e V = ****
 
Kev1n said:
Thanks, Do I insert values ie: A = 0.02xSin

So phase angle is 0
No. You know A and ω. Insert the given values for y and t, then you can solve for the phase.
 
Doc Al said:
No. You know A and ω. Insert the given values for y and t, then you can solve for the phase.

How does this look:

y = A sin(wt + b) - you are given x = 0.0192 when t = 0.05, A = 0.02 and you know w, so you can find a
d) y = A sin (wt + b)
A = 20 mm, w = 19.8 rad/s
Now y = 19.2 mm at t=0.05s
Substituting
19.2 = 20 Sin(19.8x0.05 + b)
Thus, Sin(0.99 + b) = 19.2/20
Sin(0.99 + b) = 0.96
0.99 + b = Sin-10.96 = 1.287 ( Take the angle in radians)
So b = 1.287 – 0.99 = 0.297
Or y = 20 Sin(19.8t + 0.297)

e) Similarly, v =396 Cos(19.8t + 0.297)

f) a = --7840.8 Sin(19.8t + 0.297)
 
(a) Write the expression for the displacement of the mass as a function of time.

x = Asin (wt + 0)

(b) Write the expression for the velocity of the mass as a function of time.

V = dx/dt = Aw cos (wt + 0)


(c) Write the expression for the acceleration of the mass as a function of time.

a = dv/dt = -Aw2 sin (wt + 0)
 

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